Twisted-Austere Submanifolds in Euclidean Space
A twisted-austere -fold ( , ) in ℝⁿ consists of a -dimensional submanifold of ℝⁿ together with a closed 1-form on , such that the second fundamental form A of and the 1-form satisfy a particular system of coupled nonlinear second-order PDE. Given such an object, the ''twisted c...
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| Опубліковано в: : | Symmetry, Integrability and Geometry: Methods and Applications |
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| Дата: | 2021 |
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Інститут математики НАН України
2021
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| Цитувати: | Twisted-Austere Submanifolds in Euclidean Space. Thomas A. Ivey and Spiro Karigiannis. SIGMA 17 (2021), 023, 31 pages |
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Digital Library of Periodicals of National Academy of Sciences of Ukraine| _version_ | 1859747916185337856 |
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| author | Ivey, Thomas A. Karigiannis, Spiro |
| author_facet | Ivey, Thomas A. Karigiannis, Spiro |
| citation_txt | Twisted-Austere Submanifolds in Euclidean Space. Thomas A. Ivey and Spiro Karigiannis. SIGMA 17 (2021), 023, 31 pages |
| collection | DSpace DC |
| container_title | Symmetry, Integrability and Geometry: Methods and Applications |
| description | A twisted-austere -fold ( , ) in ℝⁿ consists of a -dimensional submanifold of ℝⁿ together with a closed 1-form on , such that the second fundamental form A of and the 1-form satisfy a particular system of coupled nonlinear second-order PDE. Given such an object, the ''twisted conormal bundle'' * + is a special Lagrangian submanifold of ℂⁿ. We review the twisted austere condition and give an explicit example. Then we focus on twisted austere 3-folds. We give a geometric description of all solutions when the ''base'' is a cylinder, and when is austere. Finally, we prove that, other than the case of a generalized helicoid in ℝ⁵ discovered by Bryant, there are no other possibilities for the base . This gives a complete classification of twisted-austere 3-folds in Rⁿ.
|
| first_indexed | 2026-03-15T16:57:02Z |
| format | Article |
| fulltext |
Symmetry, Integrability and Geometry: Methods and Applications SIGMA 17 (2021), 023, 31 pages
Twisted-Austere Submanifolds in Euclidean Space
Thomas A. IVEY a and Spiro KARIGIANNIS b
a) Department of Mathematics, College of Charleston, USA
E-mail: iveyt@cofc.edu
URL: http://iveyt.people.cofc.edu
b) Department of Pure Mathematics, University of Waterloo, Canada
E-mail: karigiannis@uwaterloo.ca
URL: http://www.math.uwaterloo.ca/~karigiannis
Received October 13, 2020, in final form March 02, 2021; Published online March 10, 2021
https://doi.org/10.3842/SIGMA.2021.023
Abstract. A twisted-austere k-fold (M,µ) in Rn consists of a k-dimensional submani-
fold M of Rn together with a closed 1-form µ on M , such that the second fundamental
form A of M and the 1-form µ satisfy a particular system of coupled nonlinear second order
PDE. Given such an object, the “twisted conormal bundle” N∗M+µ is a special Lagrangian
submanifold of Cn. We review the twisted-austere condition and give an explicit example.
Then we focus on twisted-austere 3-folds. We give a geometric description of all solutions
when the “base” M is a cylinder, and when M is austere. Finally, we prove that, other than
the case of a generalized helicoid in R5 discovered by Bryant, there are no other possibilities
for the base M . This gives a complete classification of twisted-austere 3-folds in Rn.
Key words: calibrated geometry; special Lagrangian submanifolds; austere submanifolds;
exterior differential systems
2020 Mathematics Subject Classification: 53B25; 53C38; 53C40; 53D12; 58A15
1 Introduction
Special Lagrangian submanifolds are a special class of n-dimensional submanifold in Cn, and
more generally in Calabi–Yau n-folds. They were introduced by Harvey–Lawson [5] and were
the first modern example of calibrated submanifolds. They are a class of minimal (vanishing
mean curvature) submanifolds characterized by a first order nonlinear PDE, and in fact are
absolutely locally volume minimizing in their homology class. Special Lagrangian submanifolds
also play a key role in mirror symmetry through the Strominger–Yau–Zaslow conjecture [12].
They have been extensively studied by many authors. An excellent reference summarizing much
of the work on special Lagrangian geometry up to the time of its publication is the textbook [10]
of Joyce on calibrated geometry.
One particular construction of special Lagrangian submanifolds in Cn first appeared in [5] and
is known as the conormal bundle construction. Given a k-dimensional submanifold M of Rn,
Harvey–Lawson showed that its conormal bundle N∗M is special Lagrangian in T ∗Rn = Cn
if and only if M is austere, which means that all the odd degree elementary symmetric poly-
nomials in the eigenvalues of the second fundamental form vanish. Note that this is in general
a fully nonlinear second order PDE on the immersion of M in Rn. The conormal bundle con-
struction was later reviewed in detail, and generalized to the exceptional holonomies G2 and
Spin(7), by Ionel–Karigiannis–Min-Oo in [8]. Austere submanifolds in Euclidean space have
been studied by several authors, including Bryant [2] and Ionel–Ivey [6, 7].
A generalization of the conormal bundle construction was introduced by Borisenko [1] and
later significantly extended by Karigiannis–Leung [11]. The idea is as follows. Let M be a k-
dimensional submanifold of Rn. Then T ∗Rn|M = N∗M ⊕ T ∗M . Let µ be a closed 1-form
mailto:iveyt@cofc.edu
http://iveyt.people.cofc.edu
mailto:karigiannis@uwaterloo.ca
http://www.math.uwaterloo.ca/~karigiannis
https://doi.org/10.3842/SIGMA.2021.023
2 T.A. Ivey and S. Karigiannis
on M . Define the “twisted conormal bundle” to be the n-dimensional submanifold N∗M + µ =
{(νp, µp) | νp ∈ N∗pM} of T ∗Rn = Cn. This is the total space of an affine bundle over M whose
fibres are affine translates of the conormal spaces, translated by the 1-form µ. In [11] it was
proved that N∗M + µ is special Lagrangian if and only if the second fundamental form A of M
in Rn and the 1-form µ satisfy a system of coupled second order fully nonlinear PDE, which we
call the twisted austere equations. This result is stated explicitly in Theorem 2.2. (Borisenko
had only considered the case when µ is exact, n = 3, and k = 2.)
Both the original construction of Harvey–Lawson and the “twisted version” of Borisenko and
Karigiannis–Leung produce examples of ruled special Lagrangian submanifolds. Joyce [9] has
also studied ruled special Lagrangian submanifolds in Cn.
We consider the case of twisted-austere pairs
(
Mk, µ
)
in Rn for k = 1, 2, 3 and any n.
The cases k = 1, 2 are trivial to classify completely. The case k = 3 is significantly more
involved. We obtain a complete classification and give a geometric description of all possibili-
ties.
Organization of the paper and summary of results. In Section 2 we review the twisted-
austere condition, and completely describe the cases k = 1 and k = 2, as well as the case when M
is totally geodesic. We also present an explicit nontrivial solution when k = 2 and n = 3, giving
a special Lagrangian submanifold in C3. The remainder of the paper is concerned with the non
totally geodesic case when k = 3.
Section 3 establishes some general results on twisted-austere pairs
(
M3, µ
)
where M is not
totally geodesic. The main result is Theorem 3.1, where we show that M is either a generalized
helicoid swept out by planes in R5, or else n is arbitrary and M is ruled by lines. Section 4
is concerned with the particular case when M is a cylinder. The main result is Theorem 4.2,
where we give a geometric characterization of this case, in terms of a minimal surface Σ in Rn+1
and a closed 1-form λ on Σ with prescribed codifferential.
Section 5 is the heart of the paper, where we comprehensively study the case in which the
base M is austere. This study breaks up naturally into two cases, called the “split case” and the
“non-split case”, characterized by algebraic properties on the covariant derivative ∇µ. Each case
then breaks up into subcases. In the split case, M can be either a cylinder, a cone, or a “twisted
cone”. The first two subcases also occur in the non-split case. The two cylinder subcases are
related to the results of Section 4. In all subcases the twisted-austere pairs
(
M3, µ
)
with austere
base M are related to geometric data on a surface Σ, being the cross-section of the cylinder
or the link of the (twisted) cone. Using this data, the 1-form µ is described explicitly.
Finally in Section 6 we outline the proof of our classification, which is Theorem 6.1. We prove
that the pairs
(
M3, µ
)
studied in the earlier sections are the only possibilities. Three appendices
follow, collecting various technical results that are used in the main body of the paper.
2 Preliminaries
In this section we review the twisted-austere condition for a pair
(
Mk, µ
)
where Mk is a k-
dimensional submanifold of Rn and µ is a smooth 1-form on M . We also discuss the cases k = 1
and k = 2 in detail, as well as the case when Mk ⊂ Rn is totally geodesic. The remainder of the
paper is concerned with the case k = 3 for Mk not totally geodesic.
Definition 2.1. Let Mk be a k-dimensional submanifold of Rn and a let µ be a smooth 1-form
on M . Define L = N∗M + µ to be the n-dimensional submanifold of T ∗Rn given by
N∗M + µ =
{
(x, ξ + µx) ∈ T ∗Rn|M |x ∈M, ξ ∈ N∗xM
}
. (2.1)
We say that (M,µ) is a twisted-austere pair if L = N∗M+µ is a special Lagrangian submanifold
inside T ∗Rn with respect to some phase. Following [11], we refer to this as the Borisenko
construction.
Twisted-Austere Submanifolds in Euclidean Space 3
It is shown in [11] that L is Lagrangian if and only if ∇µ is a symmetric tensor on M , that is
dµ = 0. The conditions under which L is special Lagrangian are more involved. In what follows,
let Aν = ν · II denote the second fundamental form of M in the normal direction ν, and let
B = ∇µ. We use the same letters to denote the matrices that represent these covariant tensors
with respect to a local orthonormal frame field e1, . . . , ek on M . (For example, Bij = B(ei, ej),
and the Lagrangian condition is equivalent to B being a symmetric matrix.)
Theorem 2.2 (Karigiannis–Leung [11]). Fix a phase angle θ ∈ [0, 2π). Let C = I + iB, and
define the cophase angle φ by
φ = θ − (n− k)
π
2
. (2.2)
Then (M,µ) is a twisted-austere pair with phase eiθ if and only if the following three conditions
all hold:
dµ = 0, (2.3)
Im
(
eiφ detC
)
= 0, (2.4)
Im
(
ijσj
(
AνC−1
))
= 0, for all ν and all j = 1, . . . , k. (2.5)
Here σj denotes the jth elementary symmetric function of the eigenvalues of a matrix, so in
particular σ1 = tr and σk = det. (See Appendix A.1 for more details.)
Remark 2.3. In [11, Theorem 2.3] the definition of φ is the negative of what we have in (2.2),
because in [11] the definition of special Lagrangian with phase eiθ meant calibrated with respect
to e−iθdz1 ∧ · · · ∧ dzn, whereas we take it to mean calibrated with respect to eiθdz1 ∧ · · · ∧ dzn,
which is standard.
Note that condition (2.5) is really a sequence of conditions for each normal direction ν,
as follows:
Re
(
σ1
(
AνC−1
))
= 0, Im
(
σ2
(
AνC−1
))
= 0, Re
(
σ3
(
AνC−1
))
= 0, . . . . (2.6)
It is useful to rewrite equation (2.5) in the extreme cases j = 1 and j = k as follows. By the
linearity of σ1 = tr, we have
2 Re
(
σ1
(
AνC−1
))
= tr
(
Aν
(
(I + iB)−1 + (I − iB)−1
))
.
But because we can diagonalize the symmetric matrix B, it is easy to see that (I + iB)−1+
(I − iB)−1 = 2
(
I +B2
)−1
. Thus we find that
2 Re
(
σ1
(
AνC−1
))
= 2 tr
(
Aν
(
I +B2
)−1)
.
Hence, the condition (2.5) in the j = 1 case can be rewritten as
tr
(
Aν
(
I +B2
)−1)
= 0 for all ν. (2.7)
Because σk = det is multiplicative, we have σk
(
AνC−1
)
= detAν detC−1. Hence, the condi-
tion (2.5) in the j = k case can be rewritten as
(detAν) Im
(
ik
detC
)
= 0 for all ν. (2.8)
The simplest case of the twisted austere condition is when Mk ⊂ Rn is totally geodesic.
4 T.A. Ivey and S. Karigiannis
Proposition 2.4. Suppose that Mk ⊂ Rn is totally geodesic and complete. Without loss
of generality we can take Mk = Rk ⊂ Rn. Then the Borisenko construction yields a product
K × Rn−k, where K ⊂ T ∗Rk is a special Lagrangian submanifold which is the graph of µ.
Proof. Since Rk is totally geodesic, we have Aν = 0 for all ν. Thus the sequence of condi-
tions (2.6) are trivially satisfied. We have N∗M = Rk × Rn−k ⊂ Rn × Rn = T ∗Rn. The closed
1-form µ is necessarily exact, so µ = df for some f ∈ C∞
(
Rk
)
. Equation (2.4) becomes
Im
(
eiφ det(I + i Hess f)
)
= 0 for φ = θ − (n − k)π2 . Then by [5, Theorem 2.3], the graph of µ
in T ∗Rk is a special Lagrangian submanifold K of T ∗Rn with phase eiθ. (See [11, Theorem 2.3]
for discussion about the phase.) Hence L = N∗Rk + µ = K × Rn−k as claimed. �
A discussion of the cases k = 1, 2 of the twisted-austere condition was given in [11, Section 2],
which included a classification for k = 1 and a partial result for k = 2. Here we complete the
classification for k = 2. For completeness, we give the details for both cases.
Proposition 2.5. Let k = 1. If
(
M1, µ
)
is a twisted-austere pair in Rn with M complete, then
L = N∗M + µ is an n-plane in T ∗Rn = Cn.
Proof. In this case, M1 is a curve. Equation (2.3) is vacuous. The 1× 1 matrix C is 1− id∗µ.
Hence equation (2.4) becomes
sinφ = cosφ d∗µ. (2.9)
(There is a harmless sign error here in [11].) Using (2.7) for (2.5) in the j = 1 case (which is
the only allowed value of j here), and since Aν is a scalar, we get
Aν = 0 for all ν.
Thus M1 is totally geodesic, hence a straight line. Without loss of generality, we take it to be
the x-axis in Rn. Since M = R, we have µ = df for some f ∈ C∞(R). Then equation (2.9) says
that f ′′(x) = − tanφ. Hence µ = (ax+ b)dx for some constants a, b, and N∗M + µ is an affine
translation of N∗M in Cn = Rn ⊕ Rn, and is thus an n-plane. �
Remark 2.6. Proposition 2.5 is consistent with Proposition 2.4, as a special Lagrangian graph
in T ∗R1 = C2 is straight line.
Proposition 2.7. Let k = 2, and let
(
M2, µ
)
be a twisted-austere pair in Rn, such that M is
not totally geodesic. Then sinφ = 0, and M is a minimal surface in Rn with µ a closed and
coclosed 1-form on M with respect to the induced metric (and hence harmonic).
Proof. In this case, M2 is a surface and now σ2 = det. From detC = det(I+ iB) = 1 + i trB−
detB, we find that (2.4) becomes
sinφ(1− detB) + cosφ(trB) = 0. (2.10)
(There is again a harmless sign error in [11, equation (2.15)].) Using (2.7) for (2.5) in the j = 1
case gives
tr
(
Aν
(
I +B2
)−1)
= 0 for all ν. (2.11)
We also have
1
detC
=
1
1 + i trB − detB
=
(1− detB)− i trB
(1− detB)2 + (trB)2
.
Twisted-Austere Submanifolds in Euclidean Space 5
From the above, using (2.8) for (2.5) in the j = k = 2 case gives
(detAν)(trB) = 0. (2.12)
Suppose that trB 6= 0, so that detAν = 0 for all ν. Fixing a particular normal direction ν,
we can choose an orthonormal frame at a point on M such that
Aν =
(
0 0
0 a22
)
, B =
(
b11 b12
b12 b22
)
.
Then we have
(
I +B2
)−1
=
1
det
(
I +B2
) ( 1 + b222 + b212 −b12(b11 + b22)
−b12(b11 + b22) 1 + b211 + b212
)
,
and thus equation (2.11) for this ν gives a22
(
1 + b211 + b212
)
= 0, hence a22 = 0 and Aν = 0 for
this ν. Therefore whenever detAν = 0, we have Aν = 0. Since this holds for all ν, we are in the
totally geodesic case which is covered by Proposition 2.4.
Therefore we can assume there exists at least one ν such that detAν 6= 0. From (2.12)
we deduce that trB = 0, so µ is coclosed with respect to the induced metric. Since µ is also
closed by (2.3), we conclude that µ is harmonic. Now choose at a point an orthonormal frame
in which B is diagonal. Since trB = 0, in such a frame we have
B =
(
λ 0
0 −λ
)
.
But then I + B2 is a positive scalar multiple of the identity, so (2.11) implies that trAν = 0
for all ν, so M2 ⊂ Rn is a minimal surface. Finally, equation (2.10) becomes (sinφ)
(
1+λ2
)
= 0,
so sinφ = 0. �
Example 2.8. We illustrate Proposition 2.7 with an explicit example when n = 3. Throughout
this example we identify vector fields and 1-forms on R3 using the Euclidean metric. Let M2
be a surface in R3 which is given by the graph of a smooth function h : Ω→ R of two variables,
where Ω is some open set in R2. It is well known that the minimal surface equation in this
case is(
1 + h2v
)
huu +
(
1 + h2u
)
hvv − 2huhvhuv = 0. (2.13)
With respect to the global frame of tangent vector fields given by v1 = (1, 0, hu) and v2 =
(0, 1, hv), the induced metric on M2 from the Euclidean metric on R3 is
g =
(
1 + h2u huhv
huhv 1 + h2v
)
and one can compute that for a function f : Ω → R, thought of as function on the Rieman-
nian manifold (M, g), and writing the coordinates on Ω ⊆ R2 as (u1, u2) = (u, v), its exterior
derivative is
df = fu
(
g11v1 + g12v2
)
+ fv
(
g21v1 + g22v2
)
=
1
det g
((
1 + h2v
)
fu − huhvfv,−huhvfu +
(
1 + h2u
)
fv, hufu + hvfv
)
, (2.14)
6 T.A. Ivey and S. Karigiannis
and its Laplacian is
∆gf =
1√
det g
∂
∂ui
(
gij
√
det g
∂f
∂uj
)
=
1(
1 + h2u + h2v
)((1 + h2v
)
fuu +
(
1 + h2u
)
fvv − 2huhvfuv
)
− 1
(det g)2
(hufu + hvfv)
((
1 + h2v
)
huu +
(
1 + h2u
)
hvv − 2huhvhuv
)
. (2.15)
Substituting (2.13) into (2.15) eliminates the second term. We deduce that f is a harmonic
function on the minimal surface M if and only if(
1 + h2v
)
fuu +
(
1 + h2u
)
fvv − 2huhvfuv = 0. (2.16)
Using the Euclidean metric to identify covectors with tangent vectors, the conormal space is
spanned by ν∗ = (−hu,−hv, 1), and we obtain from (2.1) and (2.14) that the twisted conormal
bundle N∗M + df is identified with the submanifold
{(x1(t, u, v), x2(t, u, v), x3(t, u, v), y1(t, u, v), y2(t, u, v), y3(t, u, v)) : (u, v) ∈ Ω, t ∈ R}
in R6 ∼= C3, with coordinate functions given by
x1 = u, x2 = v, x3 = h(u, v),
y1 = −thu +
1
1 + h2u + h2v
((
1 + h2v
)
fu − huhvfv
)
,
y2 = −thv +
1
1 + h2u + h2v
(
−huhvfu +
(
1 + h2u
)
fv
)
,
y3 = t+
1
1 + h2u + h2v
(hufu + hvfv).
Proposition 2.7 says that if the two functions h and f satisfy the pair of equations (2.13)
and (2.16), then the immersion of the open set Ω×R in R2×R is a special Lagrangian submanifold
of C3 with phase ei
π
2 .
Note that in particular, if we choose f = h then the pair of equations (2.13) and (2.16)
coincide. For example, we can take h(u, v) = arctan v
u , so that M is a helicoid in R3, which is
a minimal surface. Then taking f = h, one can compute that (y1, y2, y3) is(
v
(
t
(
1 + u2 + v2
)
−
(
u2 + v2
))(
u2 + v2
)(
1 + u2 + v2
) ,−
u
(
t(1 + u2 + v2)−
(
u2 + v2
))(
u2 + v2
)(
1 + u2 + v2
) ,
1 + t
(
1 + u2 + v2
)(
1 + u2 + v2
) )
.
The authors verified directly that the above is a special Lagrangian submanifold of C3 with
phase ei
π
2 . Of course, even over the helicoid, there are infinitely many more solutions. Given
h(u, v) = arctan v
u , a computation on Maple shows that the general solution to (2.16) is
f = A1
(
arctan
v
u
+
1
2
arcsin
(
1 + 2u2 + 2v2
))
+A2
(
arctan
v
u
− 1
2
arcsin
(
1 + 2u2 + 2v2
))
,
where A1, A2 are arbitrary C2 functions of one variable. The solution f = h = arctan v
u
corresponds to A1(s) = A2(s) = 1
2s.
Twisted-Austere Submanifolds in Euclidean Space 7
3 Twisted-austere 3-folds
Because the special Lagrangian n-folds for M totally geodesic arise by taking products of lower-
dimensional examples with a flat factor, we generally exclude the case whereM is totally geodesic
from now on.
In this section we state and prove the first of our two main theorems, which characterizes
a twisted-austere pair
(
M3, µ
)
when M is a 3-dimensional submanifold of Rn that is not totally
geodesic. There are only two possibilities.
Theorem 3.1. Let (M,µ) be a twisted-austere pair where M3 ⊂ Rn is not totally geodesic, and
let φ be as in (2.2) with k = 3. Then cosφ 6= 0, and either
(i) n is arbitrary and M is ruled by lines, or else
(ii) n = 5 and M is a generalized helicoid swept out by planes in R5.
The proof of Theorem 3.1 takes up this entire section, and we break it up into a sequence
of propositions, all of which share the assumptions of Theorem 3.1.
Proposition 3.2. We have det(Aν) = 0 for all normal directions ν, and moreover cosφ 6= 0.
Proof. Recall from Theorem 2.2 that, in addition to dµ = 0 which just says that B is symmetric,
the twisted-austere conditions for 3-dimensional M are
Im
(
eiφ detC
)
= 0, (3.1)
Re
(
σ1
(
AνC−1
))
= 0, (3.2)
Im
(
σ2
(
AνC−1
))
= 0, (3.3)
Re
(
σ3
(
AνC−1
))
= 0, (3.4)
where C = I + iB. Here σ3 is the determinant. Note that detC 6= 0. (See the proof of Pro-
position A.2.)
Using Re detC = 1−σ2(B) and Im detC = σ1(B)−σ3(B), the first condition (3.1) expands as
(1− σ2(B)) sinφ+ (σ1(B)− σ3(B)) cosφ = 0. (3.5)
Next, note that we can expand conditions (3.2) and (3.3) using the identities
σ1
(
AνC−1
)
=
σ1(A
ν(I − adjB)) + 2i{Aν , B}
detC
, (3.6)
σ2
(
AνC−1
)
=
σ2(A
ν) + iσ1(B adjAν)
detC
, (3.7)
where { , } denotes the symmetric bilinear form corresponding to σ2 on the space S3 of 3 × 3
symmetric matrices (that is, {W,W} = σ2(W ) for all W ∈ S3). A general version (for k × k
matrices) of the identity (3.7) is proved in Proposition A.2 and the identity (3.6) is proved
in Proposition A.5.
Suppose that det(Aν) 6= 0. We will derive a contradiction. The last condition (3.4) implies
that
Re detC = 1− σ2(B) = 0. (3.8)
By (3.8) detC is purely imaginary, thus substituting (3.6) into (3.2) implies that {Aν , B} = 0,
while substituting (3.7) into (3.3) implies that σ2(A
ν) = 0. Together with (3.8), these in
turn imply that t 7→ B + tAν parametrizes a line on the quadric hypersurface in S3 defined
8 T.A. Ivey and S. Karigiannis
by σ2(W ) = 1. However, by Remark A.4, the signature of σ2 on S3 is (1, 5). It is well-known
(and easy to check) that this implies that the hypersurface contains no lines. Hence Aν = 0,
which contradicts our assumption that detAν 6= 0.
Now suppose that cosφ = 0. Then (3.5) implies again that Re detC = 1− σ2(B) = 0, so as
before we conclude that Aν = 0, contradicting our assumption that M is not totally geodesic. �
For use below, we note that multiplying the numerator and denominator of the right-hand
side of (3.6) by eiφ, and using the fact that by (3.1) the denominator is now real, we see that (3.2)
is equivalent to
σ1(A
ν(I − adjB)) cosφ− 2{Aν , B} sinφ = 0. (3.9)
Similarly, assuming (3.1) shows that condition (3.3) is equivalent to
σ2(A
ν) sinφ+ σ1(B adjAν) cosφ = 0. (3.10)
Lemma 3.3. The second fundamental form Aν cannot have rank one for any normal direction ν.
Proof. Suppose Aν has rank one for some ν. We will obtain a contradiction. There is a frame
with respect to which Aν = A0 is of the form
A0 =
1 0 0
0 0 0
0 0 0
.
This form is invariant under rotating the vectors e2 and e3 within the plane they span, so we
may also assume without loss of generality that B12 = 0.
Substituting Aν = A0 into (3.9) gives
(B22 +B33) sinφ+
(
B22B33 −B2
23 − 1
)
cosφ = 0. (3.11)
Equation (3.1) in this case becomes((
1 +B2
23 −B22B33
)
B11 +B2
13B22 +B22 +B33
)
cosφ
−
(
(B22 +B33)B11 −B2
13 +B22B33 −B2
23 − 1
)
sinφ = 0. (3.12)
Multiplying (3.11) by (B11 + tanφ) and adding this to (3.12) yields, after some manipulation,
that
B2
13(sinφ+B22 cosφ) + (B22 +B33) secφ = 0.
Solving this equation for B33 and substituting back into (3.11) gives(
B2
13(sinφ+B22 cosφ)2 +B2
22 +B2
23 + 1
)
cosφ = 0,
which, since cosφ 6= 0, has no real solutions. �
Proposition 3.4. At each point p ∈M , there exists an orthonormal frame with respect to which
the span
| IIp | = {ν · II | ∀ ν ∈ NpM} ⊆ S2T ∗pM
lies in one of the following subspaces:
(i) W1 =
∗ ∗ 0
∗ ∗ 0
0 0 0
, (ii) W2 =
0 0 ∗
0 0 ∗
∗ ∗ ∗
.
Moreover, if dim | IIp | = 1, then we are necessarily in case (i).
Twisted-Austere Submanifolds in Euclidean Space 9
Proof. The two possible forms for | II | follow from Proposition A.7 in Appendix A.2, and the
final statement is established in the first paragraph of the proof of Proposition A.7. �
Proposition 3.5. If M falls into case (i) of Proposition 3.4, then B33 = − tanφ with respect
to the same orthonormal frame. If M does not fall into case (i) then M is a generalized helicoid
in R5.
Proof. Suppose that we are in case (i). Then one can compute that equation (3.10) factors as(
Aν11A
ν
22 − (Aν12)
2
)
(sinφ+B33 cosφ) = 0.
Since M is not totally geodesic, Lemma 3.3 tells us that there is an Aν that has rank two.
It follows that B33 = − tanφ.
Now suppose that we are not in case (i). By Proposition 3.4 we know that dim | IIp | ≥ 2.
Also, by Lemma 3.3 we know that IIp cannot contain any rank one matrices, so it must be
two-dimensional and spanned by matrices of the form
A1 =
0 0 1
0 0 0
1 0 ∗
, A2 =
0 0 0
0 0 1
0 1 ∗
.
Substituting Aν = A1 and Aν = A2 into (3.10) yields respectively B22 = − tanφ and
B11 = − tanφ. Using these values and substituting Aν = A1 + A2 into (3.10) yields B12 = 0.
Finally using these values for B11, B12, B22 and substituting either Aν = A1 or Aν = A2
into (3.9) yields that the (3, 3) entry of (Aν) is zero, which implies that trAν = 0 for all ν. Thus
in fact M is minimal, and with respect to an appropriate basis, we have
| IIp | ⊂ V ′2 =
0 0 ∗
0 0 ∗
∗ ∗ 0
.
It now follows that | II | is simple in the sense of Bryant [2], and hence by [2, Theorem 3.1] that M
must be a generalized helicoid. Because | IIp | has dimension at most two, the first osculating
space of M at each point has dimension at most five. Moreover, because the first prolongation
of V ′2 has dimension zero it follows from Theorem A.8 in Appendix A.3 that the first osculating
space of M is fixed, so that M lies in a 5-dimensional subspace of Rn. �
Proposition 3.6. If M falls into case (i) of Proposition 3.4, then M is ruled by lines.
The proof of this proposition is relatively simple, but uses the method of moving frames.
Before giving the proof, we recall some details about the frame bundle which will be needed
in the proof as well as in later sections.
Let F be the oriented orthonormal frame bundle of Rn, whose fiber at a point p consists of all
oriented orthonormal bases of TpRn. We may think of a point u in the fiber as a matrix U ∈
SO(n) whose columns comprise the corresponding frame. The frame bundle carries a canonical
Rn-valued 1-form ω such that
ωu(v) = U−1π∗v, (3.13)
where π : F → Rn is the basepoint map and we identify π∗v ∈ Tπ(u)Rn with a column vector
in Rn in the usual way. (In other words, the entries of ωu(v) give the coefficients of the expansion
of π∗v in terms of the frame corresponding to u.) In what follows let ωr denote the components
of ω, where 1 ≤ r, s, t ≤ n.
10 T.A. Ivey and S. Karigiannis
Suppose Mk ⊂ Rn is a submanifold and f is a local section of F|M , that is a local oriented
orthonormal frame field with component vector fields e1, . . . , en. Then it follows from (3.13)
that the Rn-valued function x on M giving the position in Rn satisfies
dx = erf
∗ωr. (3.14)
In particular, if the frame f is adapted to M in the sense that e1, . . . , ek span the tangent space
to M at each point, then f∗ωa = 0 for k < a ≤ n.
The frame bundle also carries a matrix-valued connection form Ω, taking value in so(n),
which satisfies the structure equation
dω = −Ω ∧ ω, dΩ = −Ω ∧Ω.
In terms of components, these equations read
dωr = −ωrs ∧ ωs, dωrs = −ωrt ∧ ωts. (3.15)
The existence (and uniqueness) of the connection form is a special case of the existence of the
Levi-Civita connection on a Riemannian manifold N . However, when N = Rn an easy way
to obtain the connection form, in terms of its components ωrs , is to regard the members er of the
frame as Rn-valued functions on F, and resolve their exterior derivatives in terms of the frame
itself:
der = esω
s
r . (3.16)
Returning to the situation of an adapted frame field f along a submanifold Mk, it follows
from (3.16) that the pullbacks of the ωaj encode the second fundamental form of M :
II(ei, ej) =
(
ei ω̃aj
)
ea, (3.17)
where we use ei, ej for 1 ≤ i, j ≤ k to denote the frame vector fields tangent to M , and the tilde
accent denotes pullback by f .
Proof of Proposition 3.6. Let f = (e1, . . . , en) be an adapted local frame along M such that
with respect to the basis e1, e2, e3 for TpM , the space | II | assumes the form (i) in Proposition 3.4.
Then (3.17) implies that ω̃a3 = 0 for 4 ≤ a ≤ n. Then from (3.16) we have
de3 = e1ω̃
1
3 + e2ω̃
2
3. (3.18)
We will show that the frame vector e3 is tangent to a ruling along M .
By Proposition 3.4, we can assume without loss of generality that e4 · II has rank two. Then
0 = dω̃4
3 = −ω̃4
1 ∧ ω̃1
3 − ω̃4
2 ∧ ω̃2
3 = ω̃1
3 ∧
(
A1jω̃
j
)
+ ω̃2
3 ∧
(
A2jω̃
j
)
,
where Aij for 1 ≤ i, j ≤ 2 are the entries of a rank two matrix. Since the 1-forms in parentheses
on the right are linearly independent, we have
ω̃j3 ≡ 0 mod ω̃1, ω̃2. (3.19)
That is, ω̃1
3 and ω̃2
3 are linear combinations of ω̃1 and ω̃2. Then from (3.18) we have de3 ≡ 0
mod ω̃1, ω̃2. Thus, e3 is fixed as one moves along M in the direction of e3. �
Twisted-Austere Submanifolds in Euclidean Space 11
Before turning to the construction of examples of twisted-austere pairs, we now derive some
equations that relate the adapted moving frame (and the associated 1-forms) to the matri-
ces Aν , B that satisfy the twisted-austere conditions. (These equations will be needed in the
next two sections.) First, equation (3.17) can be rewritten as
ω̃ai = (Aa)ijω̃
j , (3.20)
where the matrix Aa gives the components of the second fundamental form in the direction of ea.
Next, because the ω̃i form a coframe along M , we can expand
µ = µiω̃
i.
Then ∇µ = Bijω̃
i ⊗ ω̃j , where the Bij are calculated using
dµi − µjω̃ji = Bijω̃
j . (3.21)
In terms of this equation, the results of Propositions 3.5 and 3.6 can be interpreted as follows.
For a base M carrying an adapted moving frame with respect to which | II | lies in W1, the frame
vector e3 points along the ruling. Thus, e3 µ = µ3 is a natural geometric invariant which we
will refer to as the slope of the twisted-austere pair (M,µ). (Note that this depends on a choice
of orientation for the rulings.) Then using (3.21), along with (3.19), we can interpret the
condition B33 = − tanφ as saying that the derivative of the slope along the ruling is equal
to the constant − tanφ.
4 Cylindrical examples
We saw in Theorem 3.1 that if M3 is the base of a twisted-austere pair, then either M is
ruled by lines or is a generalized helicoid in R5 which is ruled by planes. In this section we
will construct special examples of twisted-austere pairs
(
M3, µ
)
assuming that M is ruled by
parallel lines, that is M is a cylinder. From now on, it will be convenient for us to take the
ambient space as Rn+1, equipped with Euclidean coordinates x0, x1, . . . , xn such that the rulings
point in the x0 coordinate direction. Corresponding to this, we now change to using e0, e1, e2
to denote the members of the moving frame that are tangent to M , with e0 pointing along the
rulings.
Let Σ0 be the surface obtained by intersecting M with a copy of Rn perpendicular to the
rulings. (For the sake of argument, let this Rn be the hyperplane given by x0 = 0.) We can
construct an adapted moving frame along M by taking an adapted moving frame e1, e2, e3, . . . , en
along Σ0 (such that e1, e2 are tangent to the surface), parallelly translating these vectors along
the rulings, and completing the frame with the constant unit vector field e0 tangent to the rulings.
In what follows, it will be convenient to take the index ranges 0 ≤ α, β ≤ 2, 1 ≤ i, j, l,m ≤ 2
and 3 ≤ a, b ≤ n; so, for example, equation (3.21) now reads
dµα − µβω̃βα = Bαβω̃
β. (4.1)
The canonical forms and connection forms on M defined by (3.14) and (3.16) satisfy
(i) ω̃1, ω̃2 and ω̃1
2 are basic for the projection to Σ0, and the same is true for the ω̃ai ,
(ii) because e0 is constant on M , the forms ω̃i0 and ω̃a0 are zero,
(iii) as a result, the first structure equation in (3.15) implies that ω̃0 is closed.
In fact, if we let u be the restriction to M of the ambient coordinate x0, then ω̃0 = du.
12 T.A. Ivey and S. Karigiannis
Suppose that, on Σ0, we have ω̃ai = haijω̃
j , so that the haij are the components of the second
fundamental form of Σ0 as a submanifold in Rn. Pulling the ω̃ai back to M , we see that the
components of M ’s second fundamental form are given by
Aa =
(
0 0
0 haij
)
, (4.2)
where now the zeros are in the first row and column, corresponding to the tangent vector e0.
Since these matrices are singular, the highest-order twisted-austere condition (3.4) holds auto-
matically. Since M is not totally geodesic, Lemma 3.3 tells us that Aa has rank two for at least
one normal direction ea, and by Proposition 3.5 the next-highest-order twisted-austere condition
forces B00 = − tanφ.
We now consider the u-dependence of the components of µ and its covariant derivative.
Because ω̃i0 = 0 and B00 = − tanφ, equation (4.1) implies that
d(µ0 + u tanφ) = B0iω̃
i.
Since the right-hand side of the above equation is semibasic for the projection to Σ0, and recalling
that ω̃0 = du, we can write
µ0 = k secφ− u tanφ, (4.3)
where k is a smooth function on Σ0. Define the smooth functions ki on Σ0 by dk = kiω̃
i, so
that B0i = ki secφ. Then (4.1) implies that d(µi − uki secφ) is semibasic for Σ0, so we may set
µi = λi + uki secφ,
where the λi are functions on Σ0. (Note, however, that these depend on the choice of frame
on Σ0, while k does not.) Substituting these into (4.1) then gives
dλi − λjω̃ji = Bijω̃
j − u secφ
(
dki − kjω̃ji
)
.
Expanding both sides as polynomials in u and comparing coefficients, we obtain
Bij = λij + ukij secφ,
where we have set
dλi = λjω̃
j
i + λijω̃
j , dki = kjω̃
j
i + kijω̃
j .
The kij are the components of the Hessian ∇2k with respect to the coframe on Σ0, and the λij
are also symmetric in i and j, indicating that λ = λiω̃
i (which is well-defined, independent of
choice of coframe) is a closed 1-form on Σ0. In terms of these tensor components, we have
B =
(
− tanφ ki secφ
ki secφ λij + ukij secφ
)
. (4.4)
Substituting (4.2) and (4.4) into the two remaining twisted-austere conditions (3.5) and (3.9),
and equating powers of u, gives(
1 + k22
)
λ11 − 2k1k2λ12 +
(
1 + k21
)
λ22 = −
(
k21 + k22
)
tanφ, (4.5)(
1 + k22
)
k11 − 2k1k2k12 +
(
1 + k21
)
k22 = 0, (4.6)(
1 + k22
)
ha11 − 2k1k2h
a
12 +
(
1 + k21
)
ha22 = 0. (4.7)
We now give a geometric interpretation of the last two equations.
Twisted-Austere Submanifolds in Euclidean Space 13
Proposition 4.1. Let Σ0 ⊂ Rn be a surface and let k be a smooth function on Σ0. We endow Σ0
with the metric g0 it inherits from Rn. Let dk and ∇2k have components ki and kij respectively,
and let haij be the components of the second fundamental form of Σ0, relative to an adapted
orthonormal frame e1, e2, ea. Let Σ =
{
(p, k(p)) ∈ Rn × R | p ∈ Σ0
}
be the graph of k. Then Σ
is a minimal surface in Rn+1 if and only if k satisfies (4.6) and (4.7).
Proof. Let ĝ be the pullback to Σ0 of the ambient metric on Σ. Then
ĝij = δij + kikj . (4.8)
Let ωi be the dual 1-forms to the ei, let ωij be the connection forms for the metric g0 on Σ0,
and let ϕij denote the connection forms for the metric ĝ with respect to the same coframe.
Differentiating (4.8) yields that
ϕij − ωij =
(
ĝ−1
)i`
k`kjmω
m. (4.9)
Letting ∇̂ denote the covariant derivative with respect to ĝ, we can compute that relative to the
coframe ω1, ω2, we have(
∇̂2k
)
ij
=
1
det ĝ
kij .
It follows that equation (4.6) is equivalent to ∆ĝk = 0. On the other hand, equation (4.7) says
that the trace with respect to ĝ of the second fundamental form of Σ0 vanishes. That is, the
projection π : Σ→ Σ0 is harmonic.
In summary, the equations (4.6), (4.7) hold if and only if the coordinate functions on Σ are
harmonic (relative to ĝ), which in turn is equivalent to Σ being minimal. �
We now geometrically interpret the remaining equation (4.5). Recall that the 1-form λ = λiω̃
i
is closed. If we introduce a local potential function ` on Σ0 such that d` = λ, then using (4.9)
one computes that(
∇̂2`
)
ij
= λij −
(
ĝ−1
)lm
λlkmkij .
In particular, assuming that k satisfies (4.6), then equation (4.5) is equivalent to
∆ĝ` = −
∣∣∇̂k∣∣2
ĝ
tanφ.
Below, we will also express this condition in terms of the codifferential of λ.
Gathering together all our conclusions in this section, we have established the following result.
Here we drop the hats and just use the metric on the graph of k, referring to the graph of k
as Σ and its induced metric from Rn+1 as g.
Theorem 4.2. Assume that
(
M3, µ
)
is a twisted-austere pair, and that M ⊂ Rn+1 is ruled
by parallel lines. Then M is the union of lines passing through a minimal surface Σ ⊂ Rn+1.
Moreover, if we choose Euclidean coordinates x0, x1, . . . , xn such that the rulings point in the
x0-direction, then
µ = π∗λ+secφ d(u(π∗k))−tanφudu = π∗λ+secφ ((π∗k)du+ud(π∗k))−tanφudu, (4.10)
where u is the restriction of the x0 coordinate to M , k is the restriction of x0 to Σ, π : M → Σ
is the projection along the rulings, and λ is a closed 1-form on Σ satisfying
∗d∗λ = |∇k|2 tanφ, (4.11)
where the Hodge star and norms used are with respect to the metric on Σ. Conversely, given
a minimal surface Σ ⊂ Rn+1 which is everywhere transverse to a fixed coordinate direction ∂/∂x0,
and a 1-form λ satisfying (4.11) for k being the restriction of x0 to Σ, then the union of lines
through Σ parallel to this direction gives a 3-dimensional submanifold M which forms a twisted-
austere pair with µ given by (4.10).
14 T.A. Ivey and S. Karigiannis
5 Examples with austere bases
In this section we will determine all examples of twisted-austere pairs
(
M3, µ
)
where the
“base” M is austere but is not totally geodesic, nor a generalized helicoid. By Proposition 3.6
and our assumption that M is not a generalized helicoid, we know that M is ruled by lines.
As in the previous section we let Rn+1 be the ambient Euclidean space, and we number the
orthonormal frame vectors as (e0, e1, e2, e3, . . . en), where e0, e1, e2 are tangent to M with e0
pointing along the rulings.
The fact thatM is ruled also follows from Bryant’s classification of austere 3-folds in Euclidean
space [2], which asserts that M is either a product of a minimal surface in Rn with a line,
or a (possibly twisted) cone over a minimal surface in the Sn. (The twisted cone construction
will be reviewed below.)
For 3-dimensional submanifolds, the austere condition amounts to minimality and detAν = 0
for all normal directions ν. By Proposition 3.2 the twisted-austere conditions imply the deter-
minant condition. We will now see how the twisted-austere conditions simplify in the presence
of the minimality condition. With respect to the moving frame (adapted as described at the
start of this section), the matrices representing ∇µ and the second fundamental form in the
direction of er (for 3 ≤ r ≤ n) look like
B =
− tanφ B10 B20
B10 B11 B12
B20 B12 B22
, Ar =
0 0 0
0 Ar11 Ar12
0 Ar12 −Ar11
, (5.1)
respectively. (Note that we have incorporated the minimality condition.) Recall from the proof
of Proposition 3.5 that the twisted-austere condition (3.10) is satisfied by the constant value of
the top-left entry of B. In terms of these matrix entries, it is straightforward to compute that
the two remaining conditions (3.5) and (3.9) are equivalent respectively to the pair of equations
B11 +B22 + sinφ cosφ
(
B2
10 +B2
20
)
+ cos2 φ
(
B2
20B11 − 2B10B20B12 +B2
10B22
)
= 0 (5.2)
and (
B2
20 −B2
10
)
Ar11 − 2B10B20A
r
12 = 0. (5.3)
If B10 = B20 = 0, then conditions (5.2), (5.3) greatly simplify: the first becomes B11 +
B22 = 0 and the second condition becomes vacuous. In this case, the tensor B = ∇µ splits
as B = − tanφ
(
ω0
)2
+Bijω
iωj , which is the sum of a constant multiple of the square of the
arclength element along the ruling plus a quadratic form which restricts to be zero along the
rulings. We will refer to this as the split case, and the case where one of B10, B20 is always
nonzero as the non-split case.
5.1 The split case
We begin by defining an exterior differential system whose integral submanifolds correspond to
the adapted frame described above. In what follows, we will use index ranges 0 ≤ a, b, c, e ≤ 2,
1 ≤ i, j, k ≤ 2 and 3 ≤ r, s ≤ n.
To an adapted frame f along M we can associate a submanifold of the orthonormal frame
bundle F of Rn+1 by simply taking the image of f : M → F|M . However, if we want to characte-
rize submanifolds satisfying the austere conditions, we must introduce the components Arab of the
second fundamental form as extra variables, and take the image of (f,A) which is a submanifold
of F × T1, where T1 = S3 ⊗ Rn−2 is the space of Rn−2-valued symmetric bilinear forms on R3.
For example, if we were investigating submanifolds M3 whose second fundamental form satisfies
Twisted-Austere Submanifolds in Euclidean Space 15
certain algebraic conditions that defined a smooth subvariety N ⊂ T1, then on F×N we would
define 1-forms
ωra −Arabωb
(where the components Arab are taken as coordinate functions on T1) which, due to equa-
tion (3.20), would pull back to be zero on the image of (f,A) when M satisfies the conditions.
In our situation we want to impose conditions which also involve µ, so we need to introduce
the components of µ and ∇µ as additional variables. Accordingly, let T2 = T1 ×R3 ×S3 denote
the space where the tensor components (Arab, µa, Bab) take values, and let N ⊂ T2 be the affine
subspace defined by
B11 +B22 = 0, B00 = − tanφ, B0i = 0, Ar0a = 0, Ar11 +Ar22 = 0. (5.4)
(Thus, N has dimension 5 + 2(n − 2).) On F × N define 1-forms β, θ, Ω (taking value in R3,
Rn−2, and the space of (n− 2)× 3 matrices, respectively) as follows:
βa := −dµa + µbω
b
a +Babω
b, (5.5a)
θ :=
(
ω3, . . . , ωn
)T
, (5.5b)
Ωr
a := −ωra +Arabω
b. (5.5c)
Then if
(
M3, µ
)
is a twisted-austere pair where the base is austere and of split type, the image
of (f,A, µ,B) is an integral submanifold of the Pfaffian system I generated by β, θ, Ω. Because
this integral submanifold lies over M ⊂ Rn+1, it satisfies the independence condition ω0 ∧
ω1 ∧ ω2 6= 0, and we will refer to integral submanifolds satisfying this condition as admissible.
Conversely, any admissible integral submanifold of I is generated by a moving frame along an
austere M3 ⊂ Rn+1 such that (M,µ) is a twisted-austere pair of split type.
Lemma 5.1. On any admissible integral submanifold M̂3 of I, there are functions p, q such
that
ω1
0 = pω1 + qω2,
ω2
0 = −qω1 + pω2.
(5.6)
Moreover, the corresponding submanifold M ⊂ Rn+1 is one of the following three possibilities:
(i) a product of a line with a surface in Rn when p = q = 0,
(ii) a cone over a minimal surface in Sn when q = 0 but p 6= 0, or
(iii) a twisted cone when q 6= 0.
Moreover, in case (iii), integral submanifolds only exist if sinφ = 0.
Proof. The system I is algebraically generated by the component 1-forms of β, θ, Ω and their
exterior derivatives, and for M̂ to be an integral submanifold it is necessary and sufficient that
this finite list of 1-forms and 2-forms pull back to be zero on M̂ . Moreover, in computing the
generator 2-forms, any terms which are wedge products with the generator 1-forms may be omit-
ted. (This is known as computing ‘modulo the system 1-forms’, denoted by I1.) For example,
dωr = −Ωr
a ∧ ωa ≡ 0 modulo I1, and hence the exterior derivative of the components of θ do
not contribute any additional generator 2-forms to I. In the same way, using (3.15), (5.5b),
and (5.5c) we compute
dΩr
a = −dωra + dArab ∧ ωb +Arabdω
b
= ωrc ∧ ωca + ωrs ∧ ωsa + dArab ∧ ωb −Arab
(
ωbc ∧ ωc + ωbs ∧ ωs
)
≡ Arcbωb ∧ ωca + ωrs ∧Asacωc + dArab ∧ ωb −Arabωbc ∧ ωc mod I1
≡
(
dArab −Aracωcb −Arcbωca +Asabω
r
s
)
∧ ωb mod I1.
16 T.A. Ivey and S. Karigiannis
In particular, noting the zero entries in Ar from (5.4) gives dΩr
0 ≡ −Arijωi0 ∧ωj . Because Ar has
rank two for at least one r, it follows from the Cartan Lemma that on any admissible integral
submanifold we have ωi0 = P ijω
j for some functions P ij . Substituting this into the expression
for dΩr
0, we then find that in order for the 2-form dΩr
0 to vanish along M̂ , we must have
ArikP
k
j = ArjkP
k
i (5.7)
for all i, j. In other words, if we think of the P ij as entries in a 2× 2 matrix, then AP must be
symmetric whenever A is the lower-right block of a matrix Ar.
Suppose first that | II | is 2-dimensional on an open subset of M . Then the span of the lower-
right blocks of the Ar includes
(
1 0
0 −1
)
and ( 0 1
1 0 ), and substituting these into (5.7) shows that P
must have the form P =
( p q
−q p
)
for some functions p, q, as claimed.
On the other hand, suppose that | II | is 1-dimensional on M . Then we may adapt the frame
so that, say A3 = λ
(
0 0 0
0 0 1
0 1 0
)
for some λ 6= 0, and At = 0 for t > 3. In this case, one can compute
that
dΩ3
1 ∧ ω2 − dΩ3
2 ∧ ω1 ≡ λ
(
ω2
0 ∧ ω2 − ω1
0 ∧ ω1
)
∧ ω0
modulo the 1-forms in I. The left hand side of this expression is in I, and thus the right hand
side must vanish when pulled back to M̂ . Substituting ωi0 = P ijω
j into the right hand side gives
P 1
2 = −P 2
1 . Using (5.7) shows that P 1
1 = P 2
2 . Thus P indeed has the desired form.
From (3.16), and using (5.5c) and (5.4), the differential of the unit vector parallel to the
ruling on M is
de0 ≡ e1ω
1
0 + e2ω
2
0 mod I,
= p
(
e1ω
1 + e2ω
2
)
+ q
(
e1ω
2 − e2ω
1
)
using (5.6).
Comparing this with the differential of the position vector on M , given by (3.14), we see that
de0 − p dx ≡ q
(
e1ω
2 − e2ω
1
)
mod I, ω0.
In other words, if we follow a curve on M orthogonal to the rulings, the differential of the ruling
direction e0 is proportional to the differential of the position if and only if q is identically zero,
and the ruling direction is constant if and only if p and q are both identically zero. Hence, M is
a cylinder iff p = q = 0 and is a cone iff q = 0 and p 6= 0. That the remaining case, where q is
nonvanishing, corresponds to M being a twisted cone follows from Bryant’s classification [2].
Additional generator 2-forms for I are obtained by differentiating (5.5a) and using the three
equations in (5.5). One computes that
dβa ≡
(
dBab −Bacωcb −Bcbωca
)
∧ ωb + µbA
r
bcA
r
aeω
c ∧ ωe mod I1. (5.8)
In particular, from (5.1) we obtain that
dβ0 ≡ tanφω0
i ∧ ωi −Bijωi0 ∧ ωj .
Substituting (5.6) into the above shows that this 2-form equals −2q tanφω1 ∧ω2. Thus, admis-
sible integral manifolds with q 6= 0 exist only if tanφ = 0. �
We now consider the three sub-cases given by Lemma 5.1. In what follows, we will let
m = e0 µ = µ0 denote the slope, so that
µ = mω0 + µiω
i.
Note that from the form of β0 in (5.5a) we have
dm ≡ µiωi0 − tanφω0 +B0iω
i mod I1.
Twisted-Austere Submanifolds in Euclidean Space 17
5.1.1 M is a cylinder
In this case, we may write M = Σ × R where Σ is a minimal surface in Rn. We let t be the
coordinate on the R-factor, and write dt to denote the pullback to M of its differential, which
coincides with the dual ω0 of the frame vector e0. Since the ωi0 = 0 on M̂ , equation (5.5a) gives
β0 = −(dm+tanφ dt), and the vanishing of this 1-form implies that m+t tanφ is constant. Fur-
thermore, wedging (5.5a) with ωa and summing over a gives d
(
µ1ω
1 +µ2ω
2
)
≡ 0 modulo β1, β2,
which shows that µ̌ = µ − mdt is a well-defined closed 1-form on Σ, and it is easy to check
that µ̌ is harmonic. The converse also holds:
Theorem 5.2. Let Σ is an arbitrary minimal surface in Rn, µ̌ a harmonic 1-form on Σ and
let m be a linear function with derivative − tanφ. Then the cylinder M = Σ×R ⊂ Rn+1, together
with µ = µ̌+m(t) dt where t is the coordinate on the second factor, forms a twisted-austere pair.
Proof. This is a special case of the construction in Theorem 4.2 with k constant. �
5.1.2 M is a cone
In this case we assume that q = 0 identically. We will show in Remark 5.3 below that p is
nowhere zero, and thus up to a change of orientation we can assume that p > 0 everywhere.
To analyze this case, we first construct a partial prolongation of the system I, introducing the
components of ω0
1, ω0
2 and dp as new variables. To this end, let p be a coordinate on the last
factor in F×N × R+, and on this space define 1-forms
α1 := ω0
1 + pω1, α2 := ω0
2 + pω2. (5.9)
One can compute that
dα1 ≡
(
dp+ p2ω0
)
∧ ω1, dα2 ≡
(
dp+ p2ω0
)
∧ ω2
modulo I, α1, α2, and thus
dp = −p2ω0 (5.10)
on any admissible integral manifold.
Remark 5.3. It follows from (5.10) and the equations (3.14), (3.16) that the vector x− (1/p)e0
is constant, giving the position of the vertex of the cone. Equation (5.10) also implies that p
is constant along surfaces orthogonal to the rulings, whereas along the rulings it behaves like
solutions to the separable ODE dy/ds = −y2, for which 1/y is a linear function of s. Therefore p
cannot vanish, but it can blow up to infinity, which happens at the vertex of the cone.
Let α3 := dp + p2ω0 and let I+ be the Pfaffian system generated by α = (α1, α2, α3), β, θ
and Ω.
Lemma 5.4. Admissible integral manifolds of I+ exist only if tanφ = 0.
Proof. Using (5.8) and the identity B11 + B22 = 0 which holds in the split case, a lengthy
computation gives
dβ1 ∧ ω2 − dβ2 ∧ ω1 ≡ 2p tanφω0 ∧ ω1 ∧ ω2 mod I+1 . �
Using s = 1/p to denote the function on M giving the distance to the vertex of the cone, and
recalling from Lemma 5.4 that because we must have tanφ = 0, one can compute using (5.5a)
that
µ ≡ d(ms) mod I+.
18 T.A. Ivey and S. Karigiannis
Using (5.5a) and (5.6), we obtain dm = µ0ω
i
0 = p
(
µ1ω
1 +µ2ω
2
)
on solutions. In particular, dm
has no ω0 component, so the slope m is constant along the rulings, and is thus a well-defined
function on Σ.
Theorem 5.5. The slope satisfies ∆m = −2m, where ∆ denotes the Laplacian on Σ. Con-
versely, if Σ is an arbitrary minimal surface in Sn and m is a smooth function on Σ satisfying
∆m = −2m, then the cone over Σ together with µ = d(ms) is a twisted-austere pair for tanφ = 0.
Proof. Modulo the 1-forms of I+ we can compute that
dm ≡ p
(
µ1ω
1 + µ2ω
2
)
,
dµ1 − µ2ω2
1 ≡ (B11 −mp)ω1 +B12ω
2,
dµ2 + µ1ω
2
1 ≡ B12ω
1 + (B22 −mp)ω2,
Setting p = 1 to restrict to Σ, we compute using the above and (3.15) that
∗∆m = d ∗ dm = d
(
µ1ω
2 − µ2ω1
)
= dµ1 ∧ ω2 + µ1dω
2 − dµ2 ∧ ω1 − µ2dω1
=
(
µ2ω
2
1 +B11ω
1 −mω1 +B12ω
2
)
∧ ω2 − µ1ω2
i ∧ ωi
−
(
−µ1ω2
1 +B12ω
1 +B22ω
2 −mω2
)
∧ ω1 + µ2ω
1
i ∧ ωi
= (B11 +B22 − 2m)ω1 ∧ ω2.
Using B11 + B22 = 0 from (5.4), and taking Hodge star of the above, we conclude that
∆m = −2m.
Conversely, let Σ ⊂ Sn be an arbitrary minimal surface which is carrying a moving frame
(y, v1, v2, v3, . . . , vn) where the unit vector y represents position on the surface, v1, v2 are tangent
to Σ, and the vr are tangent to Sn but normal to the surface. To this moving frame we associate
canonical forms η1, η2 and connection forms η12 and ηri , such that
dy = viη
i, dvi = vjη
j
i + vrη
r
i
as Rn+1-valued functions. Because Σ is minimal, ηri = Hr
ijη
j for some traceless 2×2 matrices Hr.
Define a mapping ψ : Σ× R+ → F (with s as coordinate on the R+ factor) by
x = sy, e0 = y, e1 = v1, e2 = v2, er = vr.
This gives a moving frame along the cone over Σ.
By differentiating x and e0, e1, e2, er using (3.14) and (3.16), we compute that ψ∗ωi = sηi,
ψ∗ω0 = ds, ψ∗ωi0 = ηi, ψ∗ωri = ηri and ψ∗ωr0 = 0. It then follows that the components of the
second fundamental form of the cone, relative to ω0, ω1, ω2, are
Ar = s−1
(
0 0
0 Hr
)
.
Let the components mi, mij of the covariant derivatives of m on Σ be defined by dm = miη
i
and dmi −mjη
j
i = mijη
j . Then taking µ = d(ms) and using (5.5a) yields that the components
of ∇µ relative to ω0, ω1, ω2 are
B = s−1
0 0 0
0 m11 +m m12
0 m12 m22 +m
.
It is then easy to see that ∆m = −2m implies that (5.2) is satisfied. �
Remark 5.6. The minimality of Σ ⊂ Sn is equivalent to its coordinates as a submanifold
of Rn+1 being eigenfunctions of ∆ for eigenvalue −2. However, if we take m to be one of
these coordinate functions, the special Lagrangian submanifold in T ∗Rn+1 that results from the
Borisenko construction is easily seen to be merely a translation of the conormal bundle of M .
Twisted-Austere Submanifolds in Euclidean Space 19
5.1.3 M is a twisted cone
Recall from [2] that the twisted cone over a minimal surface in the sphere is constructed as
follows. Let u : Σ → Sn be a minimal immersion of a surface Σ, and let f be a scalar function
on Σ satisfying ∆f = −2f . (This, of course, is the same equation satisfied by the components
of u when u is regarded as an Rn+1-valued function.) Thus, the Rn+1-valued 1-form β =
u(∗df)− f(∗du) is closed, and the twisted cone is given by
X(s, t) = w(s) + tu(s), s ∈ Σ, t ∈ R+, (5.11)
where w : Σ→ Rn+1 satisfies dw = β. (It may be necessary to pass to the universal cover of Σ
for w to be well-defined.)
Let I be the Pfaffian system defined at the beginning of Section 5.1 and let I+ be a partial
prolongation defined on F×N×R2 by taking p, q as coordinates on the last factor and adjoining
1-forms
α1 := ω0
1 + pω1 + qω2, α2 := ω0
2 − qω1 + pω2.
These are analogous to the 1-forms defined in (5.9) but now we assume q 6= 0. Consequently,
from Lemma 5.1, it is necessary that tanφ = 0. By differentiating the above expressions, we can
compute as in Section 5.1.2 that
dp+
(
p2 − q2
)
ω0 = h1ω
1 + h2ω
2, dq + 2pqω0 = h2ω
1 − h1ω2, (5.12)
on any admissible integral manifold, for some undetermined functions h1, h2.
Let M̂ be an admissible integral manifold of I+ for which p and q are both nonvanishing.
If its base M is parametrized by an immersion (5.11), and the form of the right-hand side
indicates that the unit vector u must point along the rulings of M , and thus must coincide with
vector e0 of our adapted frame. In fact, one can check that if we choose u = e0, f = q/
(
q2 + p2
)
and t = p/
(
q2 + p2
)
, then there exists a w such that w + tu equals the position vector x on M .
Moreover, one can also compute explicitly using (5.5a) and (5.12) that, for these choices of f
and t, we have
µ = m(∗df)− f(∗dm) + d(tm), (5.13)
where the slope m is again an eigenfunction on Σ satisfying ∆m = −2m.
Conversely, we have the following:
Theorem 5.7. Let u : Σ → Sn be a minimal immersion and let M be a twisted cone over its
image Σ, defined by data (f,w). Let m : Σ→ R satisfy ∆m = −2m. Then (M,µ) is a twisted-
austere pair, where µ is given by (5.13), for tanφ = 0.
Sketch of proof. Let FS denote the orthonormal frame bundle of Sn, and let J be the Pfaffian
system on FS ×R2(n−2)+n+9 that encodes the minimal surface condition for Σ, coupled with the
equations satisfied by f , m and w. (On the second factor in the product, we use as coordinates
the two free components of the second fundamental form in each normal direction, the three
1-jet variables for f , the five free 2-jet variables for m, and the components of w.) Let I++
be the Pfaffian system on F × R2(n−2)+9 that is a further prolongation of I+ including the
free components of the derivatives of p, q as additional variables. Both Pfaffian systems have
rank 4n− 1, and one can define a map from the underlying manifold of I++ to the underlying
manifold of J such that I++ is the pullback of J, and there is a one-to-one correspondence
between admissible integral surfaces of J and admissible integral 3-manifolds of I++. Further
details are left to the interested reader. �
20 T.A. Ivey and S. Karigiannis
5.2 The non-split case
In this subsection we assume that (M,µ) is a twisted-austere pair where the components of ∇µ
and II (relative to the adapted frame described at the beginning of Section 5) take the form (5.1)
with B2
10 +B2
20 > 0 at every point.
Remark 5.8. In this section we are restricting to the open set U , where B2
10 + B2
20 > 0.
Since M is austere, it is minimal, and hence real analytic. Moreover, in all cases in this section,
the 1-form µ is also real analytic, as it is defined using solutions to a Laplace equation with real
analytic right hand side. It follows that U must in fact be a dense open set.
By rotating the frame vectors e1, e2 we may arrange that B20 = 0 and B10 > 0 at every
point. It then follows from (5.3) and (5.1) that the diagonal entries of Ar must all vanish, and
hence | II | is 1-dimensional at each point. Therefore, we may adapt the normal frame so that
Ar = 0 for all r > 3.
Proposition 5.9. Let
(
M3, µ
)
be a twisted-austere pair where M ⊂ Rn is austere but is not
a generalized helicoid, and such that ∇µ does not split. Then M lies in R4.
Proof. It is easy to check that the first prolongation of | II | has dimension zero, so this follows
from Theorem A.8 in Appendix A.3. �
For the rest of this section we will assume that M lies in R4, and will continue to use indices
0, . . . , 3 to label the members of the moving frame. We will let I denote the Pfaffian exterior
differential system associated to our adapted frame, analogous to that defined at the beginning
of Section 5.1, but with a much shorter list ω3, βa, Ωa of 1-form generators. Here, βa is as
defined in (5.5a), where B is now assumed to have the form
B =
− tanφ B10 0
B10 B11 B12
0 B12 B22
subject to the condition (5.2), which now takes the form
B11 +B22 +B2
10 cosφ(sinφ+ cosφB22) = 0, (5.14)
and Ωa = −ω3
a +Aabω
b, where
A =
0 0 0
0 0 h
0 h 0
.
The system I thus has rank 7, and is defined on F × N where N ⊂ R4 × S3 has coordinates
h, µa, Bab satisfying B00 = − tanφ, B20 = 0 and (5.14). (We will solve this equation for B11
in terms of the other coordinates, and assume that h and B10 are positive. By Remark 5.8 this
assumption will hold on a dense open set.)
Lemma 5.10. The conclusions of Lemma 5.1 apply in the non-split case as well, but solutions
with q 6= 0 are not possible.
Proof. The first assertion (5.6) follows by the same argument made in the proof of Lemma 5.1
for the case where | II | is 1-dimensional, and the correspondence between the values of p, q and
the branches of the Bryant classification is the same. To eliminate the possibility of twisted
Twisted-Austere Submanifolds in Euclidean Space 21
cones, we compute the system 2-forms and, using the values given by (5.6), one can compute
that
dβ0 ∧ ω1 ≡ −B10
(
2qω0 + ω2
1
)
∧ ω1 ∧ ω2
dΩ1 ∧ ω2 ≡ −h
(
qω0 + 2ω2
1
)
∧ ω1 ∧ ω2
}
mod I1.
Linearly combining the 3-forms on the right-hand sides above to eliminate ω2
1 ∧ ω1 ∧ ω2 shows
that B10hqω
1 ∧ ω2 ∧ ω3 is in the ideal. Given our assumptions that B10 and h are positive,
we see that admissible integral manifolds with q 6= 0 are not possible. �
We now consider the two sub-cases given by Lemma 5.10. As before, m = e0 µ = µ0
is the slope.
5.2.1 M is a cylinder
As in Section 5.1.1, M = Σ0 × R where Σ0 is minimal surface in R3, and we let t denote the
coordinate on the second factor, hence ω0 = dt. Using (5.6) with p = q = 0 shows that
dm ≡ − tanφω0 +B10ω
1 mod I1. (5.15)
Therefore, we see that now m+ t tanφ is non-constant, in contrast to Section 5.1.1. Neverthe-
less, M is still described by the construction of Theorem 4.2 for ambient space R4, but now
it follows from (4.3) that the function k = (m + t tanφ) cosφ is non-constant. Hence, not only
is M a product of the minimal surface Σ0 ⊂ R3 with a line, it is also the family of parallel lines
through a non-trivial minimal surface Σ ⊂ R4 which is a graph over Σ0.
The set of such pairs (Σ0,Σ) forms a 5-parameter family (modulo rigid motions) and can be
determined by solving a system of ordinary differential equations whose solutions are expressible
using elliptic functions. Since equation (5.15) shows that m is constant along the asymptotic
directions of M that are annihilated by ω2, the set of twisted-austere pairs (M,µ) where M3
is an austere cylinder with m + t tanφ non-constant is in one-to-one correspondence with the
4-parameter sub-family of these pairs (Σ0,Σ) where the minimal graph Σ has constant height
along one set of asymptotic lines of Σ0.
5.2.2 M is a cone
As in Section 5.1.2, in this case we assume that p > 0 and q = 0 identically, and we begin
by defining a partial prolongation. As before, we use p as a new coordinate and define 1-forms
α1 := ω0
1 + pω1, α2 := ω0
2 + pω2, α3 := dp+ p2ω0.
We can compute that dΩ0 ≡ 0 and
dΩ1 ≡ −2hω2
1 ∧ ω1 +
(
dh+ hpω0
)
∧ ω2,
dΩ2 ≡
(
dh+ hpω0
)
∧ ω1 + 2hω2
1 ∧ ω2
modulo I1, α1, α2. It follows from the above equations that on any admissible integral manifold,
there will be functions u1, u2 such that
ω2
1 = u1ω
1 + u2ω
2, dh = h
(
−2u2ω
1 + 2u1ω
2 − pω0
)
.
Because of this, we will thus define the prolongation I+1 on F×N×R3, with coordinates p, u1, u2
on the last factor, by adjoining α1, α2, α3 as well as
α4 := −ω2
1 + u1ω
1 + u2ω
2,
α5 := −dh+ h
(
−2u2ω
1 + 2u1ω
2 − pω0
)
.
22 T.A. Ivey and S. Karigiannis
Lemma 5.11. Austere bases of cone type with µ non-split only exist for tanφ = 0. (As in
Section 5.1.2, this means that the slope must be constant along the rulings.)
Proof. The computations are quite involved, but we carefully describe all the steps so that the
reader will be able to fill in all the details if desired.
We will compute the 2-forms modulo the newly-added 1-forms of I+. Thus, all the congru-
ences in this proof will be modulo I+1 . To begin, one can compute that
dβ0 ≡
(
dB10 + 2B10pω
0 −B10u1ω
2
)
∧ ω1.
It follows that on any admissible integral manifold there is a smooth function Z such that
dB10 = −2B10pω
0 + Zω1 +B10u1ω
2. (5.16)
On the other hand, we can compute that
dβ1 ≡ dB10 ∧ ω0 − cos2 φ
(
2B10(B22 + tanφ)dB10 +
(
B2
10 + sec2 φ
)
dB22
)
∧ ω1
+ dB12 ∧ ω2 + · · · ,
dβ2 ≡ dB12 ∧ ω1 + dB22 ∧ ω2 + · · · ,
where for the moment we have omitted terms that are “torsion” (that is, linear combinations
of ω0 ∧ ω1, ω0 ∧ ω2 and ω1 ∧ ω2). These terms come into play when we linearly combine these
2-forms in I+ so as to eliminate the terms involving dB12 and dB22, obtaining the following
3-form:
dβ1 ∧ ω2−
(
1+cos2 φB2
10
)
dβ2 ∧ ω1 ≡ −dB10 ∧
(
ω2 ∧ ω0+2 cos2 φB10(B22+tanφ)ω1 ∧ ω2
)
+
(
2p tanφ−B10u2 − cos2 φB3
10u2
)
ω0 ∧ ω1 ∧ ω2.
Substituting in for dB10 from (5.16), and solving for Z so that the 3-form on the right vanishes,
we obtain
dB10 =
[
2p tanφ−B10u2 + cos2 φB2
10(4p(B22 + tanφ)−B10u2)
]
ω1
+B10
(
u1ω
2 − 2pω0
)
. (5.17)
Next, differentiating the right-hand side of (5.17), and using the value of dB10 given by (5.17),
yields a 2-form Υ that must vanish on all integral submanifolds. Wedging this with ω2 and
linearly combining this with other 3-forms in I+ yields
sec2 φ
(
Υ− B10
2h
dα5
)
∧ ω2 +
(
4pB2
10dβ2 −B3
10dα4
)
∧ ω1
≡ 4p
(
2B3
10u2 − 3B2
10p(B22 + tanφ) + p tanφ sec2 φ
)
ω0 ∧ ω1 ∧ ω2.
Thus, all admissible integral submanifolds must lie in the zero locus of the polynomial
S1 := 2B3
10u2 − 3B2
10p(B22 + tanφ) + p tanφ sec2 φ.
We then differentiate the above expression and again use (5.17) to compute
dS1 ∧ ω1 ∧ ω2 + 3 sec2 φ
(
Υ− B10
2h
dα5
)
∧ ω2 + 5B3
10dα4 ∧ ω1
≡ 4
(
−11pB3
10u2 + 9B2
10p
2(B22 + tanφ) + 2p2 tanφ sec2 φ
)
ω0 ∧ ω1 ∧ ω2.
Twisted-Austere Submanifolds in Euclidean Space 23
This yields a second polynomial integrability condition, and eliminating B22 between the two
polynomials shows that all integral submanifolds must lie in the zero locus of
S2 := B3
10u2 − p tanφ sec2 φ.
Differentiating the above expression and again using (5.17), we obtain that
dS2 ∧ ω1 ∧ ω2 +B3
10dα4 ∧ ω1 ≡ p
(
−7B3
10u2 + p tanφ sec2 φ
)
ω0 ∧ ω1 ∧ ω2.
This last polynomial cannot vanish at the same time as S2 unless tanφ = 0. �
As in Section 5.1.2, we conclude that the slope m is a well-defined function on the minimal
surface Σ inside S3, and satisfies ∆m = −2m. However, in this case m and Σ turn out not to be
arbitrary. This is because the computations in the proof of Lemma 5.11 imply that u2 = 0 and
B22 = 0 identically on all such solutions. Consequently, the system I+ simplifies. For example,
equation (5.17) along with the vanishing of α3, α4, α5 now implies that B2
10 is a constant multiple
of hp3.
After an additional prolongation step, we obtain a Frobenius system. This means that solu-
tions of this type are determined by solving systems of ODE, and so depend on finitely many
constants. While leaving the details for the interested reader, the end result is that, up to a rigid
motion, the surface Σ is the torus in S3 that is parametrized by
(t, u) 7→ [cos(t) cos(au), cos(t) sin(au), sin(t) cos(u/a), sin(t) sin(u/a)],
where a is a positive constant. (Notice that the surface is compact if a2 is rational.) The slope
function is given by
m = (c1 cos(au) + c2 sin(au)) cos(t) + (c3 cos(u/a) + c4 sin(u/a)) sin(t),
while on M , the 1-form is µ = d(ms)+ c5du. (Here, c1, . . . , c5 are arbitrary constants. However,
since m is a linear combination of the R4 coordinates of Σ, the constant c5 should be chosen
to be nonzero so that the resulting special Lagrangian submanifold is not just a translation
of N∗M .)
6 Classification results
In this section, we prove that the only examples of twisted-austere 3-folds in Euclidean space
are precisely those that we have already discussed. More precisely we have the following result.
Theorem 6.1. Let (M,µ) be a twisted-austere pair, where M3 ⊂ Rn+1 is not totally geodesic.
Then either M is austere or M is a cylinder.
Proof. As in the proof of Lemma 5.11, we describe all the steps and leave the details to the
reader.
By Proposition 3.5 we can assume that M falls into case (i) of Proposition 3.4, because
otherwise M is an austere generalized helicoid. Thus, as we did in Section 5, we may adapt
a moving frame (e0, . . . , en) along M so that ∇µ and the second fundamental form in the
direction of er (for 3 ≤ r ≤ n) are represented respectively by
B =
− tanφ v1 v2
v1 B11 B12
v2 B12 B22
, Ar =
0 0 0
0 Ar11 Ar12
0 Ar12 Ar22
.
24 T.A. Ivey and S. Karigiannis
In terms of these, the twisted-austere condition (3.9) takes the form
vT adj(ar)v + sec2 φ tr(ar) = 0, (6.1)
where vT =
[
v1 v2
]
and ar is the lower-right 2 × 2 block of Ar. If v1 = v2 = 0 identically
then (6.1) implies that M is minimal, and hence austere since det(Ar) = 0 already. Thus,
we will assume from now on that M is not minimal, and hence that one of v1, v2 is nonzero at
each point.
Because equation (6.1) is linear condition on Ar, we see that dim | II | ≤ 2 at each point. First,
we will assume that | II | is 2-dimensional on an open set in M , and we will adapt the frame, by
rotating the normal vectors, so that Ar = 0 for r > 4. (The case where | II | is 1-dimensional,
including the case where M ⊂ R4, will be discussed later.) By Lemma 3.3 we know that Ar has
rank 2 for some r. We now further adapt the frame by rotating e1, e2 so that v2 = 0 identically.
With this adaptation, equation (6.1) now reads
v21A
r
22 + sec2 φ
(
Ar11 +Ar22
)
= 0, r = 3, 4.
Thus the vectors
[
A3
11, A
3
22
]
and
[
A4
11, A
4
22
]
are linearly dependent, and we may rotate frame
vectors e3, e4 so as to arrange that the diagonal entries of A4 are zero.
We now consider the Pfaffian system for which this moving frame corresponds to an admissible
integral manifold. As in Section 5 we let F be the frame bundle of Rn+1, and on N = R11 we
take coordinates µa, v1 > 0, Bij , A
3
ij and A4
12. (We will not yet impose conditions (6.1) on the
components of A3, or impose (5.2) on the components of B.) We define the 1-forms βa, θ
r,
and Ωr
a for 0 ≤ a ≤ 2 and 3 ≤ r ≤ n as in (5.5), and let I be the Pfaffian system on F × N
generated by β, θ, and Ω.
As in the proof of Lemma 5.1, we compute that dΩr
0 ≡ −Arijωi0∧ωj mod I1. Our assumption
that M is not minimal implies that A3 has rank two, and thus by the Cartan Lemma on any
admissible integral manifold there must be functions Pij = Pji such that ωi0 = PijA
3
jkω
k. (Note
that these coefficients are different from the P ij introduced in the proof of Lemma 5.1.) Using
this, we can compute that
dΩ4
0 ≡ A4
12
(
A3
22P22 −A3
11P11
)
ω1 ∧ ω2,
and thus there must be a function p such that P11 = pA3
22 and P22 = pA3
11.
As in Sections 5.1.2 and 5.1.3, we construct a partial prolongation I+ by adjoining the 1-forms
αi := ω0
i + PijA
3
jkω
k, with P11 = pA3
22, P22 = pA3
11
defined on F×N × R2, with p and P12 as additional variables. Computing the system 2-forms
then uncovers the additional integrability condition P12 = −pA3
12. Restricting the system I+
to the submanifold where this condition holds, one computes that
de0 ≡ p
(
A3
11A
3
22 −
(
A3
12
)2)(
e1ω
1 + e2ω
2
)
mod I+,
indicating that M must be a cone over a surface in Sn if p 6= 0, or a cylinder if p = 0 identically.
Imposing the twisted-austere condition (6.1) amounts to restricting to the smooth submani-
fold where the polynomial Q0 := v21A
3
22 + sec2 φ
(
A3
11 +A3
22
)
vanishes. We can compute that
dQ0 ≡ −p
(
A3
11A
3
22 −
(
A3
12
)2)(
Q0 + 4v21A
3
22
)
ω0 mod I+, ω1, ω2.
Thus, admissible integral submanifolds lying within this locus must also have p = 0 (in which
case M is a cylinder) or v1A
3
22 = 0 (in which case Q0 = 0 implies that M is minimal).
Twisted-Austere Submanifolds in Euclidean Space 25
Now consider the case where dim | II | = 1 at each point. We will not initially assume that
v2 = 0. We again define the partial prolongation I+ by adjoining 1-forms αi := ω0
i + PijA
3
jkω
k,
but now we cannot assume any relations among the Pij other than Pij = Pji. Again, to
impose (6.1) we must restrict to the zero locus of
Q0 := v21A
3
22 + v22A
3
11 − 2v1v2A
3
12 + sec2 φ
(
A3
11 +A3
22
)
.
By computing dQk ≡ Qk+1ω
0 mod I+, ω1, ω2, we obtain additional polynomials Q1, Q2, Q3
in whose zero locus any admissible integral manifold must lie. By rotating the frame we can
arrange that v2 = 0. This simplifies the polynomials, and we find that we must have P11 = 0
or det(Pij) = 0 on the common zero locus.
Setting P11 = 0 and v2 = 0 in Q1 implies that P12 = 0, and then substituting these in Q2
implies that P22 = 0, and thus in this case M is a cylinder. Therefore if M is not a cylinder,
we must have det(Pij) = 0. This condition is invariant under rotating the vectors e1, e2, so
in this case we may arrange that P12 = P22 = 0 instead of v2 = 0. Substituting in Q2 yields
A3
11 = 0, and substituting these values into Q1 gives either A3
12 = 0 or 3v21 = sec2 φ. If A3
12 = 0
then substituting into Q0 yields that M is totally geodesic. In the remaining case we have
P12 = P22 = A3
11 = 0 and v1 has a constant value. Computing the prolongation of I+ in this
case yields additional integrability conditions that imply M must be totally geodesic. �
Remark 6.2. In light of Proposition 3.5 and the classification theorem just proved, the only
remaining possibility for twisted-austere 3-folds other than those discussed in Sections 4 and 5
is that M is a generalized helicoid in R5. However, in this case the only possible values for the
1-form µ produce, via the Borisenko construction, a special Lagrangian submanifold in R10 that
is a translation of the conormal bundle of M by a constant vector.
A Appendix
In this appendix, we collect some linear algebraic results that are needed in the main body
of the paper. These include two identities relating the elementary symmetric polynomials with
the operation of taking the adjugate matrix, as well as some results on the spans of singular
symmetric matrices.
A.1 Identities relating σj and adj
In this section we prove the two fundamental identities (3.6) and (3.7) that are crucially used
in our classification. We prove a more general result than (3.7) valid for any k, whereas for (3.6)
we restrict to the case k = 3.
We first recall some basic facts about the elementary symmetric polynomials and the adjugate
matrix, to fix notation. Let A be a k×k matrix with complex entries. For j = 0, . . . , k we define
the jth elementary symmetric polynomial σj(A) of A by the expression
det(I + tA) =
k∑
j=0
tjσj(A). (A.1)
It is clear from (A.1) that σj
(
P−1AP
)
= σj(A) for all j. In particular we have σ0(A) = 1,
σ1(A) = trA, and σk(A) = detA. Moreover, each σj is a homogeneous polynomial of degree j
in the entries of A, so σj(λA) = λjσj(A) for all λ ∈ C.
26 T.A. Ivey and S. Karigiannis
Suppose that A is invertible. Then we can compute
det
(
I + tA−1
)
= det
(
tA−1
(
I + t−1A
))
= tk(detA)−1
k∑
j=0
(
t−1
)j
σj(A)
=
1
(detA)
k∑
j=0
tk−jσj(A) =
1
(detA)
k∑
j=0
tjσk−j(A).
We deduce from the above and (A.1) that
σj
(
A−1
)
=
1
detA
σk−j(A). (A.2)
The adjugate matrix adjA is the unique k × k matrix satisfying
(adjA)A = A(adjA) = (detA)I.
It is clear that adj
(
P−1AP
)
= P−1(adjA)P . Moreover, adjA is homogeneous of order k − 1,
so adj(λA) = λk−1 adjA for all λ ∈ C.
If A is invertible then adjA = (detA)A−1. We can use (A.2) and the homogeneity of σj
to compute that σj(adjA) = σj
(
(detA)A−1
)
= (detA)jσj
(
A−1
)
= (detA)j−1σk−j(A). By the
density of invertible matrices we conclude that
σj(adjA) = (detA)j−1σk−j(A) for all A. (A.3)
Note that the above is well-defined for all A even when j = 0, in which case it just says 1 = 1.
Lemma A.1. Let A and C be k × k complex matrices with C invertible. Then we have
σj
(
AC−1
)
= (detA)j+1−k(detC)−1σk−j(C adjA).
Proof. Assume first that A is invertible. Using (A.2), we compute
σj
(
AC−1
)
= σj
((
CA−1
)−1)
=
1
det
(
CA−1
)σk−j(CA−1)
=
detA
detC
σk−j
(
(detA)−1C adjA
)
=
detA
detC
(detA)−(k−j)σk−j(C adjA)
= (detA)j+1−k(detC)−1σk−j(C adjA),
as claimed. The result follows for all matrices A by the density of invertible matrices. �
Proposition A.2. Let A and B be k × k real matrices, and assume that B is symmetric. Let
C = I + iB. Then C is invertible and we have
σk−1
(
AC−1
)
=
σk−1(A) + iσ1(B adjA)
detC
. (A.4)
Proof. The result (A.4) we seek to prove is similarity invariant, so we can assume by the
spectral theorem that B is diagonal with real eigenvalues λ1, . . . , λk. But then C = I + iB is
diagonal with nonzero eigenvalues 1+iλ1, . . . , 1+iλk, and hence invertible. Applying Lemma A.1
with j = k − 1 gives
σk−1
(
AC−1
)
= (detC)−1σ1(C adjA).
But σ1 = tr is linear, so σ1(C adjA) = σ1((I + iB) adjA) = σ1(adjA) + iσ1(B adjA). The proof
is completed upon noting that σ1(adjA) = σk−1(A) from (A.3). �
Twisted-Austere Submanifolds in Euclidean Space 27
The next result is used to establish the second fundamental identity of this section.
Lemma A.3. Let B be a symmetric 3× 3 real matrix and let z ∈ C. Then we have
adj(I + zB) = I + z(σ1(B)I −B) + z2 adjB. (A.5)
Proof. By similarity invariance, we can assume that B is diagonal with real entries. We com-
pute explicitly that
B =
λ 0 0
0 µ 0
0 0 ν
, adjB =
µν 0 0
0 λν 0
0 0 λµ
,
I + zB =
1 + zλ 0 0
0 1 + zµ 0
0 0 1 + zν
,
adj(I + zB) =
(1 + zµ)(1 + zν) 0 0
0 (1 + zλ)(1 + zν) 0
0 0 (1 + zλ)(1 + zµ)
.
Then (A.5) can be directly verified. For example, the (1, 1) entry gives
(1 + zµ)(1 + zν) = 1 + z((λ+ µ+ ν)− λ) + z2µν,
which is clearly true. �
Before we can state the final result of this section, we need to introduce some more notation.
It is well-known that
σ2(A) =
1
2
(σ1(A))2 − 1
2
σ1
(
A2
)
.
This is the simplest of Newton’s identities. It can be verified directly for a diagonal matrix, which
implies the general case because the diagonalizable matrices are dense. By homogeneity, σ2 is
a quadratic form on the space of matrices, and by polarization we obtain an induced symmetric
bilinear form, which we denote by {·, ·}. Explicitly,
2{A,B} = σ1(A)σ1(B)− σ1(AB). (A.6)
Remark A.4. The positive-definite Frobenius norm 〈·, ·〉 on real matrices is given by 〈A,B〉 =
σ1
(
ATB
)
= tr
(
ATB
)
=
∑
i,j AijBij . Note that trA = 〈A, I〉. Thus the traceless symmetric
matrices are orthogonal to the identity matrix I with respect to 〈·, ·〉. Let A, B be symmetric.
We can write A = 1
k (trA)I + A0 where A0 is traceless and similarly for B. Then using (A.6)
we have
2{A,B} = (trA)(trB)− 〈A,B〉 = (trA)(trB)− 1
k2
(trA)(trB)〈I, I〉 − 〈A0, B0〉
=
k − 1
k
(trA)(trB)− 〈A0, B0〉.
The above computation shows that for k > 1 the symmetric bilinear form {·, ·} is a Lorentzian
inner product on the space of symmetric k×k real matrices, that is, with signature
(
1, k(k+1)
2 −1
)
.
This fact is used several times in this paper.
28 T.A. Ivey and S. Karigiannis
Proposition A.5. Let A and B be 3 × 3 real matrices, and assume that A is invertible and
that B is symmetric. Let C = I + iB. Then C is invertible and we have
σ1
(
AC−1
)
=
σ1(A(I − adjB)) + 2i{A,B}
detC
.
Proof. The invertibility of C was proved in Proposition A.2. Applying Lemma A.3 with z = i
gives
adjC = adj(I + iB) = I + iσ1(B)I − iB − adjB.
Multiplying both sides on the left by A and writing adjC = (detC)C−1 gives
(detC)AC−1 = A+ iσ1(B)A− iAB −A adjB.
Taking σ1 = tr of both sides and using linearity gives
(detC)σ1
(
AC−1
)
= σ1(A−A adjB) + i(σ1(A)σ1(B)− σ1(AB)).
Using (A.6) completes the proof. �
A.2 Spans of singular symmetric matrices
Let Sn denote the space of n× n symmetric matrices with real entries, and let Dn ⊂ Sn be the
affine variety of symmetric matrices with vanishing determinant. We determine the maximal
linear subspaces of Dn up to O(n)-conjugation, for n = 2 and n = 3.
Proposition A.6. Let W ⊂ D2 be a maximal linear subspace. Then dim W = 1 and is O(2)-
conjugate to the span of ( 1 0
0 0 ).
Proof. On the space S2, the determinant is a quadratic form with signature (1, 2), and thus D2
contains no linear subspaces of dimension greater than one. The result follows by diagona-
lization. �
Proposition A.7. Let W ⊂ D3 be a maximal linear subspace. Then W is 3-dimensional and
is O(3)-conjugate to one of
W1 =
∗ ∗ 0
∗ ∗ 0
0 0 0
, W2 =
∗ ∗ ∗∗ 0 0
∗ 0 0
.
Proof. Let V ⊂ D3 be an arbitrary linear subspace. If dim V = 1 then by diagonalization V
is conjugate to a subspace of W1. Thus we can assume dim V ≥ 2.
Case one: Suppose V contains a rank one matrix A0. By O(3)-conjugation we can assume
that
A0 =
1 0 0
0 0 0
0 0 0
.
Let V ′ = {B ∈ V |B11 = 0}. Then for any B ∈ V ′, by expansion along the top row, we have
0 = det(B + tA0) =
(
B22B33 −B2
23
)
t+ detB ∀ t ∈ R. (A.7)
Let p : V ′ → S2 denote the linear projection that gives the lower-right 2 × 2 block. Then from
the vanishing of the leading coefficient in (A.7), we see that p(V ′) ⊂ D2. By Proposition A.6,
Twisted-Austere Submanifolds in Euclidean Space 29
we know dim p(V ′) ≤ 1. If p(V ′) = {0} then V ⊆ W2. If not then we can assume by conjugation
that B23 = B33 = 0 for all B ∈ V ′. Now equation (A.7) reads 0 = B22B
2
13. If B22 = 0 for all
B ∈ V ′ then V = {A0}R ⊕ V ′ ⊂ W2. If not, then V ′ contains a matrix with B13 = 0 and hence
by scaling a matrix B0 of the form
B0 =
0 λ 0
λ 1 0
0 0 0
, λ ∈ R.
In this case, let V ′′ = ker p = {B ∈ V ′ |B22 = 0}. Then for any B ∈ V ′′ we can compute that
0 = det(B + tB0) = −B2
13t, ∀ t ∈ R.
Thus, B13 = 0 for all B ∈ V ′′, so V ′′ ⊂ W1 and V = {A0}R ⊕ {B0}R ⊕ V ′′ ⊂ W1.
Case two: Suppose that V contains no rank one matrices. Thus every nonzero matrix in V
has exactly two nonzero eigenvalues and one zero eigenvalue. Hence, the space V does not
intersect the cone defined by the equation σ2(A) = 0 except at the origin in S3. Since σ2 is
a quadratic form on S3 with signature (1, 5), and V has dimension at least two, the restriction
of σ2 to V will be negative on an open subset in V . If σ2 > 0 anywhere in V , then by continuity
there would be a nonzero A ∈ V such that σ2(A) = 0. But we have ruled that out in this case,
so we conclude that σ2(A) < 0 for all nonzero A ∈ V .
Now fix a rank two matrix A1 in V . Since σ2(A1) < 0, it has one positive and one negative
eigenvalue. Thus we can assume using O(3)-conjugation that A1 takes the form
A1 =
1 0 0
0 −λ2 0
0 0 0
, λ > 0.
Let V ′ = {B ∈ V |B11 = 0}. Then for B ∈ V ′ we must have
0 = det(B + tA1) = −λ2B33t
2 +
(
B22B33 −B2
23 + λ2B2
13
)
t+ det(B), ∀ t ∈ R. (A.8)
Thus, B33 = 0 and V ′ either lies in the subspace where B23 = λB13 or in the subspace where
B23 = −λB13; without loss of generality, we can assume the former.
Under the assumption B23 = λB13, we have det(B) = B2
13(2λB12 −B22). The function that
takes a matrix in V ′ to its (1, 3)-entry is a linear functional on the vector space V ′, so its kernel
is either all of V ′ or a subspace of V ′ of positive codimension. Hence, either B13 = 0 for all
matrices B in V ′, or else B13 is nonzero on a dense open subset of V ′. If B13 = 0 for all matrices
in V ′, then V ′ ⊂ W1 and hence V ⊂ W1. Otherwise, B13 6= 0 on a dense open subset of V ′,
and then det(B) = 0 from (A.8) implies that B22 = 2λB12 for all matrices in V ′. In that case,
conjugating the matrices in V ′ by the rotation matrix
R =
cos θ sin θ 0
− sin θ cos θ 0
0 0 1
,
where sin θ = λ cos θ, yields a matrix in W2. Since RA1R
−1 ∈ W2 as well, we conclude that V is
conjugate to a subspace of W2. �
A.3 A result for codimension reduction
In this section we establish a technical result that is used in the proofs of Propositions 3.5
and 5.9.
30 T.A. Ivey and S. Karigiannis
Let V and W be real vector spaces. Given a linear subspace L ⊂ SkV ∗⊗W , the prolongation
of L is defined to be
L(1) = V ∗ ⊗ L ∩ Sk+1V ∗ ⊗W.
This definition arises in the study of the tableaux associated to systems of linear first-order PDE.
See [3, Chapter VIII] for more details. For example, one can check that if V = Rn, W = R
and L is the set of symmetric n× n matrices in block form(
0 B
Bt 0
)
,
where B is an arbitrary k × (n− k) matrix, then L(1) = 0.
We use the above definition (in the special case where W = R) to formulate a codimension-
reduction theorem for submanifolds of Euclidean space.
Theorem A.8. Let Mm ⊂ RN be a smooth connected submanifold with second fundamental
form II such that the first normal bundle N1M has constant rank ρ. (Recall that the fiber at p
of N1M is the image of IIp : TpM⊗TpM → NpM.) If at each point p in M , the set | II |p as a sub-
space of S2T ∗pM satisfies | II |(1)p = 0, then M is contained in a totally geodesic submanifold R
of dimension m+ ρ which is tangent to TpM ⊕N1
pM at each p ∈M .
Proof. Near any point of M , choose an orthonormal frame e1, . . . , eN such that e1, . . . , em
span TpM and em+1, . . . , em+ρ span N1
pM . (In what follows, use index ranges 1 ≤ α, β ≤ N ,
1 ≤ i, j, k ≤ m, m < a, b ≤ m+ ρ and r, s > m+ ρ.) Let ωi, ωαβ be the canonical and connection
1-forms associated to this moving frame along M . Then
dej ≡ eahajkωk mod TpM, (A.9)
where | II |p equals the span of the symmetric matrices hajk. Suppose that
dea = erq
r
aiω
i mod TpM,N1
pM.
Then differentiating (A.9) shows that
0 = erq
r
aiω
i ∧ hajkωk.
For each r, it follows that Srijk = qraih
a
jk satisfies Srijk = Srikj , and hence belongs to the
space | II |(1), which is zero. Since the matrices hajk are linearly independent, the qrai vanish.
Hence the span of {ei, ea} is fixed. If we let R be the totally geodesic submanifold tangent
to this space at one point p ∈ M , then connecting any other point q ∈ M to p with a smooth
curve in M shows that all other points of M must lie inside R. �
A generalization of this result to submanifolds in a Riemannian manifold may be found in [4,
Section 4.2].
Acknowledgements
The authors thank the anonymous referees for useful feedback and comments that improved the
quality of the paper.
Twisted-Austere Submanifolds in Euclidean Space 31
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https://doi.org/10.1016/0550-3213(96)00434-8
https://arxiv.org/abs/hep-th/9606040
1 Introduction
2 Preliminaries
3 Twisted-austere 3-folds
4 Cylindrical examples
5 Examples with austere bases
5.1 The split case
5.1.1 M is a cylinder
5.1.2 M is a cone
5.1.3 M is a twisted cone
5.2 The non-split case
5.2.1 M is a cylinder
5.2.2 M is a cone
6 Classification results
A Appendix
A.1 Identities relating sigma j and adj
A.2 Spans of singular symmetric matrices
A.3 A result for codimension reduction
References
|
| id | nasplib_isofts_kiev_ua-123456789-211165 |
| institution | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| issn | 1815-0659 |
| language | English |
| last_indexed | 2026-03-15T16:57:02Z |
| publishDate | 2021 |
| publisher | Інститут математики НАН України |
| record_format | dspace |
| spelling | Ivey, Thomas A. Karigiannis, Spiro 2025-12-25T13:20:08Z 2021 Twisted-Austere Submanifolds in Euclidean Space. Thomas A. Ivey and Spiro Karigiannis. SIGMA 17 (2021), 023, 31 pages 1815-0659 2020 Mathematics Subject Classification: 53B25; 53C38; 53C40; 53D12; 58A15 arXiv:2006.15119 https://nasplib.isofts.kiev.ua/handle/123456789/211165 https://doi.org/10.3842/SIGMA.2021.023 A twisted-austere -fold ( , ) in ℝⁿ consists of a -dimensional submanifold of ℝⁿ together with a closed 1-form on , such that the second fundamental form A of and the 1-form satisfy a particular system of coupled nonlinear second-order PDE. Given such an object, the ''twisted conormal bundle'' * + is a special Lagrangian submanifold of ℂⁿ. We review the twisted austere condition and give an explicit example. Then we focus on twisted austere 3-folds. We give a geometric description of all solutions when the ''base'' is a cylinder, and when is austere. Finally, we prove that, other than the case of a generalized helicoid in ℝ⁵ discovered by Bryant, there are no other possibilities for the base . This gives a complete classification of twisted-austere 3-folds in Rⁿ. The authors thank the anonymous referees for useful feedback and comments that improved the quality of the paper. en Інститут математики НАН України Symmetry, Integrability and Geometry: Methods and Applications Twisted-Austere Submanifolds in Euclidean Space Article published earlier |
| spellingShingle | Twisted-Austere Submanifolds in Euclidean Space Ivey, Thomas A. Karigiannis, Spiro |
| title | Twisted-Austere Submanifolds in Euclidean Space |
| title_full | Twisted-Austere Submanifolds in Euclidean Space |
| title_fullStr | Twisted-Austere Submanifolds in Euclidean Space |
| title_full_unstemmed | Twisted-Austere Submanifolds in Euclidean Space |
| title_short | Twisted-Austere Submanifolds in Euclidean Space |
| title_sort | twisted-austere submanifolds in euclidean space |
| url | https://nasplib.isofts.kiev.ua/handle/123456789/211165 |
| work_keys_str_mv | AT iveythomasa twistedausteresubmanifoldsineuclideanspace AT karigiannisspiro twistedausteresubmanifoldsineuclideanspace |