- Email: [email protected]
Stress energy tensor of C-stationary maps
Journal of Geometry and Physics 137 (2019) 217–227
Contents lists available at ScienceDirect
Journal of Geometry and Physics journal homepage: www.elsevier.com/locate/geomphys
Stress energy tensor of C-stationary maps Nobumitsu Nakauchi Graduate School of Sciences and Technology for Innovation, Yamaguchi University, Japan
article
info
a b s t r a c t We consider a functional Econf (f ) on the space of smooth maps f between Riemannian manifolds, such that f is a weakly conformal map if and only if Econf (f ) = 0. The quantity Econf (f ) is a measure of weak conformality of maps f . Stationary maps for this functional are called C-stationary maps (Nakauchi, 2011, Kawai and Nakauchi, 2013). Weakly conformal maps are always C-stationary ones. On the other hand, in the field of calculus of variations on Riemannian manifolds, the theory of harmonic maps is developed with applications to other fields. (See two reports by Eells and Lemaire, 1978, Eells and Lemaire, 1988.) Several concepts are given in this theory. The stress energy tensor of harmonic maps is first defined by Baird and Eells (1981). This tensor is a conserved quantity. In this paper we introduce a stress energy tensor of C-stationary maps and give some results. © 2018 Elsevier B.V. All rights reserved.
Article history: Received 10 June 2018 Received in revised form 22 November 2018 Accepted 5 December 2018 Available online 19 December 2018 MSC: 58E99 58E20 53C43 Keywords: Variational problem Weakly conformal map C-stationary map Stress energy tensor
1. Introduction Let (M , g), (N , h) be Riemannian manifolds. A smooth map f from M into N is a conformal map if and only if there exists a smooth positive function ϕ on M such that f ∗ h = ϕ g, where f ∗ h denotes the pullback of the metric h by f , i.e., (f ∗ h) (X , Y ) = h df (X ), df (Y ) .
(
)
In this situation we utilize a symmetric 2-tensor Tf : = f ∗ h −
1 m
∥df ∥2 g .
where
∥df ∥2 =
∑ (
h df (ei ), df (ei ) ,
)
i
{ei } is a local orthonormal frame on (M , g), and m denotes the dimension of the manifold M. Then we can verify that f is a conformal map if and only if Tf = 0, unless df ̸ = 0. We consider a functional ∫ 2 Tf dvg , Econf (f ) = M
where dvg denotes the volume form of (M , g), and
∥Tf ∥2 =
∑
Tf (ei , ej )2 .
i, j
E-mail address: [email protected]. https://doi.org/10.1016/j.geomphys.2018.12.004 0393-0440/© 2018 Elsevier B.V. All rights reserved.
218
N. Nakauchi / Journal of Geometry and Physics 137 (2019) 217–227
The quantity Econf (f ) is a measure of weak conformality of maps f . Minimizers of Econf are close to conformal maps, even if there does not exist any conformal map from M into N. In [5] the author introduced the above functional Econf , and in [4] such a map is called C-stationary if it is a critical point of the functional Econf , i.e., if the first variation of Econf at f vanishes. Any conformal map or more generally any weakly conformal map is a C-stationary map. On the other hand, in the field of calculus of variations on Riemannian manifolds, the theory of harmonic maps is developed with applications to other fields. (See two reports by Eells and Lemaire [2], [3].) Several concepts are given in this theory. The stress energy tensor of harmonic maps is first defined by Baird and Eells [1]. This tensor is a conserved quantity. In this paper we introduce a stress energy tensor µf of C-stationary maps f . We give the following three results: (1) µf is trace-free and divergence-free (Proposition 1). (2) µf = 0 if and only if f is a weakly conformal map (Theorem 1). (3) µf appears in a first variation formula with respect to deformations of metrics on domains (Theorem 2): µf is a ‘‘gradient’’ of the functional Econf for deformations of metrics on domains. The contents of this paper are as follows: 1. 2. 3. 4. 5.
Introduction Preliminaries Stress energy tensor of C-stationary maps Vanishing stress energy tensor Deformation of metrics and stress energy tensor of C-stationary maps
2. Preliminaries In this section we give some preliminary notions and basic results. See [5] for precise arguments. Let f be a smooth map from a Riemannian manifold (M , g) into a Riemannian manifold (N , h). In this section we give a tensor Tf of conformality for any smooth map f . We recall here the following two notions. Definition 1 (Conformal and Weakly Conformal Maps). (i) A map f is conformal if there exists a smooth positive function ϕ on M such that f ∗h = ϕg .
(1)
(ii) A map f is weakly conformal if there exists a smooth non-negative function ϕ on M satisfying (1). The condition (1) is equivalent to f ∗h =
1 m
∥df ∥2 g ,
(2)
1 since taking the trace of the both sides of (1) (with respect to the metric g), we have ∥df ∥2 = mϕ , i.e., ϕ = m ∥df ∥2 . Then f is weakly conformal if and only if it satisfies (2). Note that f is weakly conformal if and only if for any point x ∈ M, f is conformal at x or dfx = 0. Taking the above situation into consideration, we utilize the symmetric 2-tensor def
Tf = f ∗ h −
1 m
∥df ∥2 g ,
i.e., def
Tf (X , Y ) = (f ∗ h)(X , Y ) −
1 m
∥df ∥2 g(X , Y )
(3)
( ) 1 = h df (X ), df (Y ) − ∥df ∥2 g(X , Y ) . m
We can easily check the following basic properties of this tensor Tf . (See Lemma 1 in [5] or Lemma 1 in [4] for a proof of this lemma.) Lemma 1. (a) Tf is symmetric, i.e., Tf (X , Y ) = Tf (Y , X ). (b) f is weakly conformal if and only if Tf = 0. (c) Tf is trace-free (with respect to the metric g), i.e.,
N. Nakauchi / Journal of Geometry and Physics 137 (2019) 217–227
trg Tf = (g , Tf ) =
m ∑
219
g(ei , ej ) Tf (ei , ej )
i, j=1 m ∑
=
Tf (ei , ei ) = 0 .
i=1
(d) The trace of Tf with respect to the pullback f ∗ h is equal to the squared norm ∥Tf ∥2 , i.e., trf ∗ h Tf = (f ∗ h, Tf ) =
m ∑
(f ∗ h)(ei , ej )Tf (ei , ej )
i, j=1
= ∥ Tf ∥2 . 1 ∥df ∥4 . m In the above equalities, the product ( , ) denotes the pairing of the symmetric 2-tensors, i.e., (e) ∥Tf ∥2 = ∥f ∗ h∥2 −
m ∑
(A, B) =
A(ei , ej ) B(ei , ej )
i, j=1
for any symmetric 2-tensors A, B, where {ei } is a local orthonormal frame. In this paper, we are concerned with the functional of the norm of Tf
∫
∥Tf ∥2 dvg .
Econf (f ) = M
This quantity Econf (f ) gives a measure of the conformality of maps f . Note that if f is a weakly conformal map, then Econf (f ) vanishes. Take any smooth deformation F of f , i.e., any smooth map F : (−ε, ε ) × M −→ N s.t. F (0, x) = f (x), where ε is a positive constant. Let ft⏐(x) = F (t , x). Then we have f0 (x) = f (x). We often say a deformation ft (x) instead of a deformation F (t , x). Let X = dF ( ∂∂t )⏐t =0 denote the variation vector field of the deformation F . We define an f ∗ TN-valued 1-form ξf on M by
ξf (X ) =
∑
Tf (X , ej )df (ej ) ,
(4)
j
where {ej } is a local orthonormal frame. The 1-form ξf plays an important role in our arguments. The first variation of the functional Econf is given by the following formula which is proved in [5]. Lemma 2 (First Variation Formula).
⏐ ⏐ ⏐
dEconf (ft ) ⏐ dt
∫
h X , divg ξf
(
= −4
)
dvg ,
M
t =0
where divg ξf denotes the divergence of ξf , i.e., divg ξf =
m ∑
(∇ei ξf )(ei ).
i=1
We give here the notion of C-stationary maps. Definition 2 (C-stationary Map). A smooth map f is called a C-stationary map if the first variation of Econf (f ) at f identically vanishes, i.e.,
⏐ ⏐ ⏐
dEconf (ft ) ⏐ dt
=0
t =0
for any smooth deformation ft of f . By the first variation formula (Lemma 2), a smooth map f is C-stationary if and only if it satisfies the Euler–Lagrange equation divg ξf = 0 , where ξf is the symmetric 2-tensor defined by (4).
(5)
220
N. Nakauchi / Journal of Geometry and Physics 137 (2019) 217–227
Obviously any weakly conformal map is a C-stationary one. There exists a C-stationary map which is not weakly conformal. See Example in [4], p. 156. 3. Stress energy tensor of C-stationary maps The stress energy tensor of harmonic maps is first defined by Baird and Eells [1]. In our case, we introduce the stress energy tensor of C-stationary maps as follows: Definition 3 (Stress Energy Tensor of C-stationary Maps). Let (M , g) and (N , h) be Riemannian manifolds. For any smooth map f from M into N, we define the stress energy tensor µf of C-stationary maps by def
µf (X , Y ) =
=
=
1 4 1 4 1 4
∥Tf ∥2 g(X , Y ) −
m ∑
Tf (X , ej ) Tf (Y , ej ) −
j=1
∥Tf ∥2 g(X , Y ) −
m ∑
1 m
∥df ∥2 Tf (X , Y )
Tf (X , ej ) (f ∗ h)(Y , ej )
j=1
∥Tf ∥2 g(X , Y ) −
m ∑
(f ∗ h)(X , ej ) Tf (Y , ej )
j=1
where {ej } is a local orthonormal frame. The above last two equalities follow from the definition of Tf . Indeed, since Tf = f ∗ h − Tf (Y , ej ) = (f ∗ h)(Y , ej ) −
1 m
1 m
∥df ∥2 g, i.e.,
∥df ∥2 g(Y , ej ) ,
we have m ∑
Tf (X , ej ) Tf (Y , ej )
j=1
=
m ∑
Tf (X , ej ) (f ∗ h)(Y , ej ) −
j=1
=
m ∑
Tf (X , ej ) (f ∗ h)(Y , ej ) −
j=1 m
=
∑
Tf (X , ej ) (f ∗ h)(Y , ej ) −
j=1
1 m 1 m 1 m
∥df ∥2
m ∑
Tf (X , ej ) g(Y , ej )
j=1 m ( ∑ ) ∥df ∥2 Tf X , g(Y , ej )ej j=1
∥df ∥2 Tf (X , Y ) .
Thus we have m ∑
Tf (X , ej ) Tf (Y , ej ) +
j=1
1 m
∥df ∥2 Tf (X , Y ) =
m ∑
Tf (X , ej ) (f ∗ h)(Y , ej ) .
(6)
j=1
Similarly we see (or directly from the symmetry with respect to X and Y ) m ∑
Tf (X , ej ) Tf (Y , ej ) +
j=1
1 m
∥df ∥2 Tf (X , Y ) =
m ∑
(f ∗ h)(X , ej ) Tf (Y , ej ) .
(7)
j=1
These two relations (6) and (7) give the last two equalities in Definition 3. Note that the stress energy tensor is defined for not only C-stationary maps but also general smooth maps. The stress energy tensor µf is a symmetric 2-tensor. To prove that it is divergence-free for C-stationary maps, we give the following fact: Lemma 3.
µf (X , Y ) = =
1 4 1 4
( ) ∥Tf ∥2 g(X , Y ) − h ξf (X ), df (Y ) ( ) ∥Tf ∥2 g(X , Y ) − h df (X ), ξf (Y ) .
(8)
N. Nakauchi / Journal of Geometry and Physics 137 (2019) 217–227
221
Proof. A simple calculation gives m ∑
m ∑
Tf (X , ej ) (f ∗ h)(Y , ej ) =
j=1
Tf (X , ej ) h df (Y ), df (ej )
(
)
(9)
j=1 m ( ) ∑ = h df (Y ), Tf (X , ej ) df (ej ) j=1 (4)
= h(df (Y ), ξf (X )) = h(ξf (X ), df (Y )) .
We have similarly (or directly from the symmetry with respect to X and Y ) m ∑
(f ∗ h)(X , ej ) Tf (Y , ej ) = h(df (X ), ξf (Y )) .
(10)
j=1
Then by Definition 3, we have Lemma 3. □ We give two basic properties: Proposition 1. m−4
trg µf =
(1) (trace)
4
∥Tf ∥2 ,
where m is the dimension of M, and trg α denotes the trace of a symmetric 2-tensor α , i.e., trg α =
m ∑
α (ei , ei ) ,
i=1
and {ej } is a local orthonormal frame. divg µf (X ) = − h divg ξf , df (X ) ,
(
(2) (divergence)
)
(
)
where divg α denotes the divergence of a symmetric 2-tensor α with respect to the first component, i.e., divg α =
m ∑
(∇ei α )(ei , X ) .
i=1
A direct consequence of this proposition is as follows: Corollary 1. (a) trg µf = 0 in the case that M is 4-dimensional. (b) divg µf = 0 for any C-stationary map f . Proof of Proposition 1. (a) trg µf
=
m ∑
µf (ei , ei )
i=1 Definition 3
=
Lemma 1(d)
=
=
1 4 m
∥Tf ∥2
m ∑
g(ei , ei ) −
i=1
m m ∑ ∑
Tf (ei , ej ) (f ∗ h)(ei , ej )
i=1 j=1
2
2
∥Tf ∥ − ∥Tf ∥ 4 m−4 ∥Tf ∥2 . 4
(b) We first note the equality m ∑ j, k=1
(
Tf (ej , ek ) ∇X Tf (ej , ek )
)
=
m ∑ j, k=1
( (
Tf (ej , ek ) ∇X h df (ej ), df (ek )
))
.
(11)
222
N. Nakauchi / Journal of Geometry and Physics 137 (2019) 217–227
Indeed we have m ∑
(
Tf (ej , ek ) ∇X Tf (ej , ek )
)
j, k=1
=
m ∑
Tf (ej , ek ) ∇X
{
1
h df (ej ), df (ek ) −
(
)
m
j, k=1
=
m ∑
Tf (ej , ek ) ∇X h df (ej ), df (ek )
( (
))
−
j, k=1
=
m ∑
Tf (ej , ek ) ∇X h df (ej ), df (ek )
( (
∥df ∥2 g(ej , ek )
m 1 ∑
m
}
Tf (ej , ek ) ∇X ∥df ∥2 g (ej , ek )
(
)
j, k=1
))
j, k=1 m 1 ∑
−
m
Tf (ej , ek ) ∇X (∥df ∥2 )g(ej , ek ) −
j, k=1
m
∇ g =0
=
∑
Tf (ej , ek ) ∇X h df (ej ), df (ek )
( (
))
−
j, k=1 Lemma 1(c)
=
m ∑
Tf (ej , ek ) ∇X h df (ej ), df (ek )
( (
))
1 m
m 1 ∑
m
Tf (ej , ek ) ∥df ∥2 (∇X g)(ej , ek )
j, k=1 m ∑
∇X (∥df ∥2 )
Tf (ej , ek ) g(ej , ek )
j, k=1
.
j, k=1
Then we have m ∑
(∇ek µf )(ek , X )
(divg µf )(X ) =
k=1 Lemma 3
=
m 1∑
4
m ∑ ( ) ( ) ∇ek ∥Tf ∥2 g (ek , X ) − h (∇ek ξf )(ek ), df (X )
k=1
k=1 m
−
∑ ( ) h ξf (ek ), (∇ek df )(X ) k=1
=
m m ∑ ) ( ) 1 1 ∑( ∇ek ∥Tf ∥2 g(ek , X ) + ∥Tf ∥2 ∇ek g (ek , X )
4
4
k=1
−h
m (∑
)
(∇ek ξf )(ek ), df (X )
−
k=1 ∇ g =0 definition of ξf
=
1 4
∇∑ m
k=1
−
k=1 m
∑ ( ) h ξf (ek ), (∇ek df )(X ) k=1
g(X , ek )ek
( ) ∥Tf ∥2 − h divg ξf , df (X )
m m ∑ ∑
Tf (ej , ek ) h df (ej ), (∇ek df )(X )
(
)
j=1 k=1
=
1 4
( ) ∇X ∥Tf ∥2 − h divg ξf , df (X ) −
m ∑
Tf (ej , ek ) h df (ej ), (∇X df )(ek )
(
)
j, k=1
( ∵ a basic property of the Hessian: (∇X df )(Y ) = (∇Y df )(X ) ) symmetry of Tf
=
1 4
m ( ( ( ) )) 1 ∑ ∇X ∥Tf ∥2 − h divg ξf , df (X ) − Tf (ej , ek )∇X h df (ej ), df (ek )
2
j, k=1
N. Nakauchi / Journal of Geometry and Physics 137 (2019) 217–227
1
(11)
=
4
m ( ) ( ) 1 ∑ ∇X ∥Tf ∥2 − h divg ξf , df (X ) − Tf (ej , ek )∇X Tf (ej , ek )
2
j, k=1
⎛ 1
=
4
223
∇X ∥Tf ∥2 − h divg ξf , df (X ) −
(
1
)
1 4
∇X ⎝
m ∑
⎞ Tf (ej , ek )2 ⎠
j, k=1
1
∇X ∥Tf ∥2 − h divg ξf , df (X ) − ∇X ∥Tf ∥2 4 ( ) − h divg ξf , df (X ) . □
=
(
)
4
=
4. Vanishing stress energy tensor In this section we prove the following result. Theorem 1. The following two facts are equivalent: (1) µf = 0 (2) f is a weakly conformal map. Proof. (1) ⇒ (2): Suppose µf = 0. By Definition 3, we see m ∑
1
Tf (X , ek ) (f ∗ h)(Y , ek ) =
4
k=1
∥Tf ∥2 g(X , Y )
Let X = ei and let Y = ej and then we have m ∑
Tf (ei , ek ) (f ∗ h)(ej , ek ) =
k=1
1 4
∥Tf ∥2 δij ,
(12)
where δij denotes the Kronecker’s delta:
δij =
{
1 0
if i = j if i ̸ = j.
We consider two symmetric matrices A, B: def
(
Tf (ei , ej ) ; i, j = 1, . . . , m
def
(
(f ∗ h)(ei , ej ) ; i, j = 1, . . . , m
A = B =
) )
.
Then (12) implies AB = BA =
1 4
∥Tf ∥2 E ,
where E denotes the identity matrix. Take any point P ∈ M. Suppose, on the contrary, that Tf ̸ = 0 at P. Then
˜ AB = B˜ A = E
(13)
at P, where def ˜ A =
4
∥Tf ∥2
A.
Since B is symmetric and positive semi-definite, it can be diagonalized, i.e., there exists an orthogonal matrix R such that
⎛ ⎜
RBR−1 = ⎜ ⎝
λ1
⎞ λ2
..
.
⎟ ⎟ ⎠ λm
224
N. Nakauchi / Journal of Geometry and Physics 137 (2019) 217–227
where λ1 , . . . , λm ≥ 0. Since A = B −
RAR−1
1 (tr B) E, m
⎛ m 1 ∑ λj λ − 1 ⎜ m ⎜ j=1 ⎜ ⎜ ⎜ ⎜ = ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝
and hence RAR−1 = RBR−1 −
1 tr (RBR−1 ) E, m
we have
⎞ m
λ2 −
1 ∑ m
λj
j=1
..
. λm
⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟. ⎟ ⎟ ⎟ ⎟ m ⎟ 1 ∑ ⎠ λj − m
(14)
j=1
On the other hand, ˜ A is positive semi-definite and therefore so is A, since by (13), ˜ A is the inverse matrix of the symmetric, positive semi-definite matrix B. Then (14) leads us to get
λi −
m 1 ∑
m
λj ≥ 0
(15)
j=1
for any i. Let λi0 be a minimum of them, i.e., λi0 ≤ λi for any i. Then we have
λi0 =
1(
) ) (15) 1( λi0 + · · · + λi0 ≤ λ1 + · · · + λm ≤ λi0 . m m m
Then all the above inequalities are equalities, and hence λi = λi0 for any i. Therefore f ∗ h = λi0 g at P. Take the trace of 1 the both sides of this equality, and then we have ∥df ∥2 = λi0 m, i.e., λi0 = m ∥df ∥2 . Then f ∗ h = m1 ∥df ∥2 g at P. This is a contradiction to the assumption Tf ̸ = 0 at P. Thus we conclude Tf = 0 everywhere, i.e., f is a weakly conformal map. (2) ⇒ (1): If f is a weakly conformal map, then Tf = 0, therefore µf = 0 by the definition of µf . □ 5. Deformations of metrics and stress energy tensor In this section we prove that the stress energy tensor is a ‘‘gradient’’ of the functional Econf for deformations of metrics on the domain M. By moving the metric g, our functional Econf (f ) can be regarded as a functional of the metric g: Econf (f , g) =
∫
∥Tf ∥2g dvg , M
where the norm ∥Tf ∥ depends on the metric g and then we denote it by the notation ∥Tf ∥g . Theorem 2. Let gt be a 1-parameter family of metrics on M satisfying g0 = g. Then the functional Econf has the following first variation with respect to the metric g:
⏐ ⏐ Econf (f , gt ) ⏐⏐ dt
∫
d
( ) µf , w dvg ,
= 2 t =0
M
where
⏐ ⏐ ∂ ⏐⏐ ⏐ w = gt ⏐ = ∇ ∂ gt ⏐ . ∂t t =0 ∂ t t =0 def
Proof. We have d
⏐ ⏐
Econf (f , gt ) ⏐⏐ dt
(16) t =0
⏐ ∫ ⏐ d 2 = ∥Tf ∥g dvgt ⏐⏐ t dt t =0 ⏐ ) ⏐ ) ( ∫ (M ∫ ⏐ ⏐ ∂ ∂ 2 ⏐ = ∥Tf ∥g ⏐ dvgt + ∥Tf ∥2g dvgt ⏐⏐ t ∂t ∂t M M t =0 t =0
N. Nakauchi / Journal of Geometry and Physics 137 (2019) 217–227
225
Using (6), (7), (9) and (10), we define a tensor γ by
γ (X , Y ) =
m ∑
1
Tf (X , ej ) Tf (Y , ej ) +
m
j=1
=
m ∑
∥df ∥2 Tf (X , Y )
(17)
Tf (X , ej ) (f ∗ h)(Y , ej ) = h ξf (X ), df (Y )
(
)
j=1
=
m ∑
(f ∗ h)(X , ej ) Tf (Y , ej ) = h df (X ), ξf (Y ) .
(
)
j=1
Note that the stress energy tensor µf = 41 ∥Tf ∥2 g − γ . We use a local frame {ei } which is orthonormal with respect to the metric g = g0 and fix it. Then the tensor gt is represented by matrix (gt )ij with respect to the frame ei : (gt )ij = gt (ei , ej ) . Note that, when t = 0, gij = g(ei , ej ) = δij , where δij denotes the Kronecker’s delta. By Lemma 1(d), we have
∥Tf ∥2g = t
(
where (gt )
ij
m ∑
(gt )ik (gt )jℓ Tf (ei , ej ) (f ∗ h)(ek , eℓ ) ,
i, j, k, ℓ=1
)
(
)
is the inverse matrix of (gt )ij . Note
m ∑ ∂ ∂ (gt )ik = − (gt )ip (gt )pq (gt )qk ∂t ∂t p, q=1
since we differentiate both sides of the inverse relation m ∑
(gt )ip (gt )pq = δqi ,
p=1
where δqi denotes the Kronecker’s delta, i.e.,
δ = i q
{
1 0
if i = q if i ̸ = q
to have m m ∑ ∑ ∂ ∂ (gt )ip (gt )pq + (gt )ip (gt )pq = 0 ∂t ∂t p=1
p=1
and multiplying the inverse matrix (gt )qk , we have the above equality (18). Then we see
⏐ ⏐ ⏐ ik jℓ ∗ (gt ) (gt ) Tf (ei , ej ) (f h)(ek , eℓ ) ⏐⏐ ⏐ i, j, k, ℓ=1 t =0 ⏐ m ∑ ∂ ⏐ 2 (g )ik ⏐ g jℓ Tf (ei , ej ) (f ∗ h)(ek , eℓ ) ∂ t t ⏐t =0 i, j, k, ℓ=1 ⏐ ( ) m ∑ ⏐ qk ip ∂ ⏐ g −2 (g )pq g g jℓ Tf (ei , ej ) (f ∗ h)(ek , eℓ ) ∂ t t ⏐t =0 i, j, k, ℓ, p, q=1 ⏐ ⏐ m m ∑ ∑ ⏐ ∂ ∗ −2 Tf (ei , ej ) (f h)(ek , ej ) (g )ik ⏐ ∂ t t ⏐⏐ i, k=1 j=1 t =0 ⏐ m ∑ ⏐ ∂ −2 γ (ei , ek ) gt (ei , ek )⏐⏐ ∂t t =0 i, k=1 ⏐ ) ( ∂ ⏐⏐ −2 γ, g , ∂ t t ⏐t =0
⏐ ⏐ ∂ ∂ 2 ⏐ ∥Tf ∥g ⏐ = t ∂t ∂t t =0 = (18)
=
=
(17)
=
=
m ∑
(18)
226
N. Nakauchi / Journal of Geometry and Physics 137 (2019) 217–227
i.e.,
⏐ ⏐ ( ) ∂ ∥Tf ∥2g ⏐⏐ = −2 γ, w . t ∂t t =0
(19)
On the other hand we use here a local coordinate (x1 , . . . , xm ) to calculate represented by the following matrix (gt )ij
( (gt )ij = gt where
∂ (i ∂ xi
dvgt
∂ ∂ , ∂ xi ∂ xj
)
∂ dvgt ⏐t =0 ∂t
⏐
locally. The metric tensor gt is
,
= 1, 2, . . . , m) is the coordinate basis of the tangent space. Note √ ( √ ) = det (gt )ij dx1 · · · dxm and dvg = det(gij ) dx1 · · · dxm .
Lemma 4 provides us with
( )⏐ ⏐ m ∑ ( )⏐ ∂ (gt )ij ⏐⏐ ∂ ij det (gt )ij ⏐⏐ = g ⏐ ⏐ ∂t ∂t t =0 i,j=1
det(gij ) . t =0
Then we have
⏐ ⏐ √ ( ⏐ )⏐ ∂ ∂ ⏐ = dx1 · · · dxm dvgt ⏐ det (gt )ij ⏐⏐ ∂t ∂t t =0 t =0 ⏐ ) ( )⏐ ( 1 ∂ 1 = √ ( ) dx1 · · · dxm det (gt )ij ⏐⏐ 2 ∂ t t =0 det gij ⎛ ( ) ⏐⏐ ⎞ m ∑ ( ) ∂ (g ) 1 1 t ij ⏐ g ij = √ ( ) ⎝ ⏐ ⎠ det gij dx1 · · · dxm ⏐ 2 ∂t det gij i,j=1 t =0 ⏐ )√ ( ⏐ ( ) 1 ∂ ⏐ = trg g det gij dx1 · · · dxm 2 ∂ t t ⏐t =0 ) 1( = trg w dvg
(20)
2 ) 1( = g , w dvg , 2
where trg
(
∂ g ⏐ ∂ t t t =0
⏐
)
denotes the trace of the 2-tensor
⏐ ⏐ d Econf (f , gt ) ⏐⏐ dt t =0 ⏐ ∫ ⏐ d 2 = ∥Tf ∥g dvgt ⏐⏐ t dt t =0 ⏐ ∫ ∫ M ⏐ ∂ ∂ 2 ⏐ ∥Tf ∥2g = ∥Tf ∥g ⏐ dvg + dvgt t ∂ t ∂ t M t =0 ∫M ( ) 1 = − 2γ + ∥Tf ∥2 g , w dv 2 M ∫ (1 ) 2 ∥Tf ∥ g − γ , w dv =2 4 ∫M ( ) µ f , w dv . □ =2
∂ g ⏐ ∂ t t t =0
⏐
with respect to g. Therefore by (16), (19) and (20), we have
⏐ ⏐ ⏐ ⏐
t =0
M
In the rest of this section, we prove the following lemma which is used in the proof of Theorem 2.
(
)
Lemma 4. For any matrix-valued function A(t) = aij (t) with symmetry aij = aji , we have
⎛ ⎞ ( ) m ∑ ∂ det A ∂ A ∂ a ij ⎠ det A , = tr A−1 det A = ⎝ aij ∂t ∂t ∂t i,j=1
where (aij ) denotes the inverse matrix of A = (aij ).
(21)
N. Nakauchi / Journal of Geometry and Physics 137 (2019) 217–227
227
Proof. We first diagonalize the matrix A, i.e., there exists an orthogonal matrix P such that
⎛
λ1
PAP −1 = ⎝
⎜
⎞ ..
.
λm
⎟ ⎠.
Then we get
∂ ∂ det A = det (PAP −1 ) ∂t ∂t ) ∂ ( λ1 . . . λm = ∂ t ( ) 1 ∂λ1 1 ∂λm = + ··· + λ1 . . . λm λ1 ∂ t λm ∂ t ) ( ∂ (PAP −1 ) det A = tr (PAP −1 )−1 ∂t
(22)
Note that P depends on t, since A is a function of t. We have
) ∂ (PAP −1 ) ∂t { }) ( ∂ A −1 ∂ P −1 ∂ P AP −1 + P P + PA PA−1 P −1 ∂t ∂t ∂t ( ( )) ( ) ( ) ∂ P ∂ A ∂ P −1 −1 −1 −1 −1 −1 (PA P ) AP + tr PA P + tr P ∂t ∂t ∂t ) ) ( ) ( ) (( ∂ A −1 ∂ P −1 ∂ P −1 (PA−1 P −1 ) + tr PA−1 + tr P AP P ∂t ∂t ∂t ( ) ( ) ( ) ∂ P −1 ∂ P −1 ∂ A P + tr P + tr PA−1 P −1 ∂t ∂t ∂t ( ) ( ) −1 ∂ (PP ) ∂A + tr A−1 ∂t ∂t ( ) ∂ A A−1 . ∂t (
tr
= tr = tr = tr = tr = tr = tr
(PAP −1 )−1
(23)
Lemma 4 follows from (22) and (23), since
( tr
∂A A−1 ∂t
)
⎛ = ⎝
m ∑ i,j=1
⎞ ∂ a ij ⎠ aij ∂t
□
Acknowledgment This work was partially supported by the Grant-in-Aid for Scientific Research (C) No. 18K03280 at Japan Society for the Promotion of Science. References [1] P. Baird, J. Eells, A conservation law for harmonic maps, in: Geometry Symposium Utrecht 1980, in: Lecture Note in Math, vol. 894, Springer-Verlag, 1981, pp. 1–25. [2] J. Eells, L. Lemaire, A report on harmonic maps, Bull. Lond. Math. Soc. 10 (1978) 1–68. [3] J. Eells, L. Lemaire, Another report on harmonic maps, Bull. Lond. Math. Soc. 20 (1988) 385–524. [4] S. Kawai, N. Nakauchi, Weak conformality of stable stationary maps for a functional related to conformality, Differential Geom. Appl. 31 (2013) 151–165. [5] N. Nakauchi, A variational problem related to conformal maps, Osaka J. Math. 48 (2011) 719–741.
Contents lists available at ScienceDirect
Journal of Geometry and Physics journal homepage: www.elsevier.com/locate/geomphys
Stress energy tensor of C-stationary maps Nobumitsu Nakauchi Graduate School of Sciences and Technology for Innovation, Yamaguchi University, Japan
article
info
a b s t r a c t We consider a functional Econf (f ) on the space of smooth maps f between Riemannian manifolds, such that f is a weakly conformal map if and only if Econf (f ) = 0. The quantity Econf (f ) is a measure of weak conformality of maps f . Stationary maps for this functional are called C-stationary maps (Nakauchi, 2011, Kawai and Nakauchi, 2013). Weakly conformal maps are always C-stationary ones. On the other hand, in the field of calculus of variations on Riemannian manifolds, the theory of harmonic maps is developed with applications to other fields. (See two reports by Eells and Lemaire, 1978, Eells and Lemaire, 1988.) Several concepts are given in this theory. The stress energy tensor of harmonic maps is first defined by Baird and Eells (1981). This tensor is a conserved quantity. In this paper we introduce a stress energy tensor of C-stationary maps and give some results. © 2018 Elsevier B.V. All rights reserved.
Article history: Received 10 June 2018 Received in revised form 22 November 2018 Accepted 5 December 2018 Available online 19 December 2018 MSC: 58E99 58E20 53C43 Keywords: Variational problem Weakly conformal map C-stationary map Stress energy tensor
1. Introduction Let (M , g), (N , h) be Riemannian manifolds. A smooth map f from M into N is a conformal map if and only if there exists a smooth positive function ϕ on M such that f ∗ h = ϕ g, where f ∗ h denotes the pullback of the metric h by f , i.e., (f ∗ h) (X , Y ) = h df (X ), df (Y ) .
(
)
In this situation we utilize a symmetric 2-tensor Tf : = f ∗ h −
1 m
∥df ∥2 g .
where
∥df ∥2 =
∑ (
h df (ei ), df (ei ) ,
)
i
{ei } is a local orthonormal frame on (M , g), and m denotes the dimension of the manifold M. Then we can verify that f is a conformal map if and only if Tf = 0, unless df ̸ = 0. We consider a functional ∫ 2 Tf dvg , Econf (f ) = M
where dvg denotes the volume form of (M , g), and
∥Tf ∥2 =
∑
Tf (ei , ej )2 .
i, j
E-mail address: [email protected]. https://doi.org/10.1016/j.geomphys.2018.12.004 0393-0440/© 2018 Elsevier B.V. All rights reserved.
218
N. Nakauchi / Journal of Geometry and Physics 137 (2019) 217–227
The quantity Econf (f ) is a measure of weak conformality of maps f . Minimizers of Econf are close to conformal maps, even if there does not exist any conformal map from M into N. In [5] the author introduced the above functional Econf , and in [4] such a map is called C-stationary if it is a critical point of the functional Econf , i.e., if the first variation of Econf at f vanishes. Any conformal map or more generally any weakly conformal map is a C-stationary map. On the other hand, in the field of calculus of variations on Riemannian manifolds, the theory of harmonic maps is developed with applications to other fields. (See two reports by Eells and Lemaire [2], [3].) Several concepts are given in this theory. The stress energy tensor of harmonic maps is first defined by Baird and Eells [1]. This tensor is a conserved quantity. In this paper we introduce a stress energy tensor µf of C-stationary maps f . We give the following three results: (1) µf is trace-free and divergence-free (Proposition 1). (2) µf = 0 if and only if f is a weakly conformal map (Theorem 1). (3) µf appears in a first variation formula with respect to deformations of metrics on domains (Theorem 2): µf is a ‘‘gradient’’ of the functional Econf for deformations of metrics on domains. The contents of this paper are as follows: 1. 2. 3. 4. 5.
Introduction Preliminaries Stress energy tensor of C-stationary maps Vanishing stress energy tensor Deformation of metrics and stress energy tensor of C-stationary maps
2. Preliminaries In this section we give some preliminary notions and basic results. See [5] for precise arguments. Let f be a smooth map from a Riemannian manifold (M , g) into a Riemannian manifold (N , h). In this section we give a tensor Tf of conformality for any smooth map f . We recall here the following two notions. Definition 1 (Conformal and Weakly Conformal Maps). (i) A map f is conformal if there exists a smooth positive function ϕ on M such that f ∗h = ϕg .
(1)
(ii) A map f is weakly conformal if there exists a smooth non-negative function ϕ on M satisfying (1). The condition (1) is equivalent to f ∗h =
1 m
∥df ∥2 g ,
(2)
1 since taking the trace of the both sides of (1) (with respect to the metric g), we have ∥df ∥2 = mϕ , i.e., ϕ = m ∥df ∥2 . Then f is weakly conformal if and only if it satisfies (2). Note that f is weakly conformal if and only if for any point x ∈ M, f is conformal at x or dfx = 0. Taking the above situation into consideration, we utilize the symmetric 2-tensor def
Tf = f ∗ h −
1 m
∥df ∥2 g ,
i.e., def
Tf (X , Y ) = (f ∗ h)(X , Y ) −
1 m
∥df ∥2 g(X , Y )
(3)
( ) 1 = h df (X ), df (Y ) − ∥df ∥2 g(X , Y ) . m
We can easily check the following basic properties of this tensor Tf . (See Lemma 1 in [5] or Lemma 1 in [4] for a proof of this lemma.) Lemma 1. (a) Tf is symmetric, i.e., Tf (X , Y ) = Tf (Y , X ). (b) f is weakly conformal if and only if Tf = 0. (c) Tf is trace-free (with respect to the metric g), i.e.,
N. Nakauchi / Journal of Geometry and Physics 137 (2019) 217–227
trg Tf = (g , Tf ) =
m ∑
219
g(ei , ej ) Tf (ei , ej )
i, j=1 m ∑
=
Tf (ei , ei ) = 0 .
i=1
(d) The trace of Tf with respect to the pullback f ∗ h is equal to the squared norm ∥Tf ∥2 , i.e., trf ∗ h Tf = (f ∗ h, Tf ) =
m ∑
(f ∗ h)(ei , ej )Tf (ei , ej )
i, j=1
= ∥ Tf ∥2 . 1 ∥df ∥4 . m In the above equalities, the product ( , ) denotes the pairing of the symmetric 2-tensors, i.e., (e) ∥Tf ∥2 = ∥f ∗ h∥2 −
m ∑
(A, B) =
A(ei , ej ) B(ei , ej )
i, j=1
for any symmetric 2-tensors A, B, where {ei } is a local orthonormal frame. In this paper, we are concerned with the functional of the norm of Tf
∫
∥Tf ∥2 dvg .
Econf (f ) = M
This quantity Econf (f ) gives a measure of the conformality of maps f . Note that if f is a weakly conformal map, then Econf (f ) vanishes. Take any smooth deformation F of f , i.e., any smooth map F : (−ε, ε ) × M −→ N s.t. F (0, x) = f (x), where ε is a positive constant. Let ft⏐(x) = F (t , x). Then we have f0 (x) = f (x). We often say a deformation ft (x) instead of a deformation F (t , x). Let X = dF ( ∂∂t )⏐t =0 denote the variation vector field of the deformation F . We define an f ∗ TN-valued 1-form ξf on M by
ξf (X ) =
∑
Tf (X , ej )df (ej ) ,
(4)
j
where {ej } is a local orthonormal frame. The 1-form ξf plays an important role in our arguments. The first variation of the functional Econf is given by the following formula which is proved in [5]. Lemma 2 (First Variation Formula).
⏐ ⏐ ⏐
dEconf (ft ) ⏐ dt
∫
h X , divg ξf
(
= −4
)
dvg ,
M
t =0
where divg ξf denotes the divergence of ξf , i.e., divg ξf =
m ∑
(∇ei ξf )(ei ).
i=1
We give here the notion of C-stationary maps. Definition 2 (C-stationary Map). A smooth map f is called a C-stationary map if the first variation of Econf (f ) at f identically vanishes, i.e.,
⏐ ⏐ ⏐
dEconf (ft ) ⏐ dt
=0
t =0
for any smooth deformation ft of f . By the first variation formula (Lemma 2), a smooth map f is C-stationary if and only if it satisfies the Euler–Lagrange equation divg ξf = 0 , where ξf is the symmetric 2-tensor defined by (4).
(5)
220
N. Nakauchi / Journal of Geometry and Physics 137 (2019) 217–227
Obviously any weakly conformal map is a C-stationary one. There exists a C-stationary map which is not weakly conformal. See Example in [4], p. 156. 3. Stress energy tensor of C-stationary maps The stress energy tensor of harmonic maps is first defined by Baird and Eells [1]. In our case, we introduce the stress energy tensor of C-stationary maps as follows: Definition 3 (Stress Energy Tensor of C-stationary Maps). Let (M , g) and (N , h) be Riemannian manifolds. For any smooth map f from M into N, we define the stress energy tensor µf of C-stationary maps by def
µf (X , Y ) =
=
=
1 4 1 4 1 4
∥Tf ∥2 g(X , Y ) −
m ∑
Tf (X , ej ) Tf (Y , ej ) −
j=1
∥Tf ∥2 g(X , Y ) −
m ∑
1 m
∥df ∥2 Tf (X , Y )
Tf (X , ej ) (f ∗ h)(Y , ej )
j=1
∥Tf ∥2 g(X , Y ) −
m ∑
(f ∗ h)(X , ej ) Tf (Y , ej )
j=1
where {ej } is a local orthonormal frame. The above last two equalities follow from the definition of Tf . Indeed, since Tf = f ∗ h − Tf (Y , ej ) = (f ∗ h)(Y , ej ) −
1 m
1 m
∥df ∥2 g, i.e.,
∥df ∥2 g(Y , ej ) ,
we have m ∑
Tf (X , ej ) Tf (Y , ej )
j=1
=
m ∑
Tf (X , ej ) (f ∗ h)(Y , ej ) −
j=1
=
m ∑
Tf (X , ej ) (f ∗ h)(Y , ej ) −
j=1 m
=
∑
Tf (X , ej ) (f ∗ h)(Y , ej ) −
j=1
1 m 1 m 1 m
∥df ∥2
m ∑
Tf (X , ej ) g(Y , ej )
j=1 m ( ∑ ) ∥df ∥2 Tf X , g(Y , ej )ej j=1
∥df ∥2 Tf (X , Y ) .
Thus we have m ∑
Tf (X , ej ) Tf (Y , ej ) +
j=1
1 m
∥df ∥2 Tf (X , Y ) =
m ∑
Tf (X , ej ) (f ∗ h)(Y , ej ) .
(6)
j=1
Similarly we see (or directly from the symmetry with respect to X and Y ) m ∑
Tf (X , ej ) Tf (Y , ej ) +
j=1
1 m
∥df ∥2 Tf (X , Y ) =
m ∑
(f ∗ h)(X , ej ) Tf (Y , ej ) .
(7)
j=1
These two relations (6) and (7) give the last two equalities in Definition 3. Note that the stress energy tensor is defined for not only C-stationary maps but also general smooth maps. The stress energy tensor µf is a symmetric 2-tensor. To prove that it is divergence-free for C-stationary maps, we give the following fact: Lemma 3.
µf (X , Y ) = =
1 4 1 4
( ) ∥Tf ∥2 g(X , Y ) − h ξf (X ), df (Y ) ( ) ∥Tf ∥2 g(X , Y ) − h df (X ), ξf (Y ) .
(8)
N. Nakauchi / Journal of Geometry and Physics 137 (2019) 217–227
221
Proof. A simple calculation gives m ∑
m ∑
Tf (X , ej ) (f ∗ h)(Y , ej ) =
j=1
Tf (X , ej ) h df (Y ), df (ej )
(
)
(9)
j=1 m ( ) ∑ = h df (Y ), Tf (X , ej ) df (ej ) j=1 (4)
= h(df (Y ), ξf (X )) = h(ξf (X ), df (Y )) .
We have similarly (or directly from the symmetry with respect to X and Y ) m ∑
(f ∗ h)(X , ej ) Tf (Y , ej ) = h(df (X ), ξf (Y )) .
(10)
j=1
Then by Definition 3, we have Lemma 3. □ We give two basic properties: Proposition 1. m−4
trg µf =
(1) (trace)
4
∥Tf ∥2 ,
where m is the dimension of M, and trg α denotes the trace of a symmetric 2-tensor α , i.e., trg α =
m ∑
α (ei , ei ) ,
i=1
and {ej } is a local orthonormal frame. divg µf (X ) = − h divg ξf , df (X ) ,
(
(2) (divergence)
)
(
)
where divg α denotes the divergence of a symmetric 2-tensor α with respect to the first component, i.e., divg α =
m ∑
(∇ei α )(ei , X ) .
i=1
A direct consequence of this proposition is as follows: Corollary 1. (a) trg µf = 0 in the case that M is 4-dimensional. (b) divg µf = 0 for any C-stationary map f . Proof of Proposition 1. (a) trg µf
=
m ∑
µf (ei , ei )
i=1 Definition 3
=
Lemma 1(d)
=
=
1 4 m
∥Tf ∥2
m ∑
g(ei , ei ) −
i=1
m m ∑ ∑
Tf (ei , ej ) (f ∗ h)(ei , ej )
i=1 j=1
2
2
∥Tf ∥ − ∥Tf ∥ 4 m−4 ∥Tf ∥2 . 4
(b) We first note the equality m ∑ j, k=1
(
Tf (ej , ek ) ∇X Tf (ej , ek )
)
=
m ∑ j, k=1
( (
Tf (ej , ek ) ∇X h df (ej ), df (ek )
))
.
(11)
222
N. Nakauchi / Journal of Geometry and Physics 137 (2019) 217–227
Indeed we have m ∑
(
Tf (ej , ek ) ∇X Tf (ej , ek )
)
j, k=1
=
m ∑
Tf (ej , ek ) ∇X
{
1
h df (ej ), df (ek ) −
(
)
m
j, k=1
=
m ∑
Tf (ej , ek ) ∇X h df (ej ), df (ek )
( (
))
−
j, k=1
=
m ∑
Tf (ej , ek ) ∇X h df (ej ), df (ek )
( (
∥df ∥2 g(ej , ek )
m 1 ∑
m
}
Tf (ej , ek ) ∇X ∥df ∥2 g (ej , ek )
(
)
j, k=1
))
j, k=1 m 1 ∑
−
m
Tf (ej , ek ) ∇X (∥df ∥2 )g(ej , ek ) −
j, k=1
m
∇ g =0
=
∑
Tf (ej , ek ) ∇X h df (ej ), df (ek )
( (
))
−
j, k=1 Lemma 1(c)
=
m ∑
Tf (ej , ek ) ∇X h df (ej ), df (ek )
( (
))
1 m
m 1 ∑
m
Tf (ej , ek ) ∥df ∥2 (∇X g)(ej , ek )
j, k=1 m ∑
∇X (∥df ∥2 )
Tf (ej , ek ) g(ej , ek )
j, k=1
.
j, k=1
Then we have m ∑
(∇ek µf )(ek , X )
(divg µf )(X ) =
k=1 Lemma 3
=
m 1∑
4
m ∑ ( ) ( ) ∇ek ∥Tf ∥2 g (ek , X ) − h (∇ek ξf )(ek ), df (X )
k=1
k=1 m
−
∑ ( ) h ξf (ek ), (∇ek df )(X ) k=1
=
m m ∑ ) ( ) 1 1 ∑( ∇ek ∥Tf ∥2 g(ek , X ) + ∥Tf ∥2 ∇ek g (ek , X )
4
4
k=1
−h
m (∑
)
(∇ek ξf )(ek ), df (X )
−
k=1 ∇ g =0 definition of ξf
=
1 4
∇∑ m
k=1
−
k=1 m
∑ ( ) h ξf (ek ), (∇ek df )(X ) k=1
g(X , ek )ek
( ) ∥Tf ∥2 − h divg ξf , df (X )
m m ∑ ∑
Tf (ej , ek ) h df (ej ), (∇ek df )(X )
(
)
j=1 k=1
=
1 4
( ) ∇X ∥Tf ∥2 − h divg ξf , df (X ) −
m ∑
Tf (ej , ek ) h df (ej ), (∇X df )(ek )
(
)
j, k=1
( ∵ a basic property of the Hessian: (∇X df )(Y ) = (∇Y df )(X ) ) symmetry of Tf
=
1 4
m ( ( ( ) )) 1 ∑ ∇X ∥Tf ∥2 − h divg ξf , df (X ) − Tf (ej , ek )∇X h df (ej ), df (ek )
2
j, k=1
N. Nakauchi / Journal of Geometry and Physics 137 (2019) 217–227
1
(11)
=
4
m ( ) ( ) 1 ∑ ∇X ∥Tf ∥2 − h divg ξf , df (X ) − Tf (ej , ek )∇X Tf (ej , ek )
2
j, k=1
⎛ 1
=
4
223
∇X ∥Tf ∥2 − h divg ξf , df (X ) −
(
1
)
1 4
∇X ⎝
m ∑
⎞ Tf (ej , ek )2 ⎠
j, k=1
1
∇X ∥Tf ∥2 − h divg ξf , df (X ) − ∇X ∥Tf ∥2 4 ( ) − h divg ξf , df (X ) . □
=
(
)
4
=
4. Vanishing stress energy tensor In this section we prove the following result. Theorem 1. The following two facts are equivalent: (1) µf = 0 (2) f is a weakly conformal map. Proof. (1) ⇒ (2): Suppose µf = 0. By Definition 3, we see m ∑
1
Tf (X , ek ) (f ∗ h)(Y , ek ) =
4
k=1
∥Tf ∥2 g(X , Y )
Let X = ei and let Y = ej and then we have m ∑
Tf (ei , ek ) (f ∗ h)(ej , ek ) =
k=1
1 4
∥Tf ∥2 δij ,
(12)
where δij denotes the Kronecker’s delta:
δij =
{
1 0
if i = j if i ̸ = j.
We consider two symmetric matrices A, B: def
(
Tf (ei , ej ) ; i, j = 1, . . . , m
def
(
(f ∗ h)(ei , ej ) ; i, j = 1, . . . , m
A = B =
) )
.
Then (12) implies AB = BA =
1 4
∥Tf ∥2 E ,
where E denotes the identity matrix. Take any point P ∈ M. Suppose, on the contrary, that Tf ̸ = 0 at P. Then
˜ AB = B˜ A = E
(13)
at P, where def ˜ A =
4
∥Tf ∥2
A.
Since B is symmetric and positive semi-definite, it can be diagonalized, i.e., there exists an orthogonal matrix R such that
⎛ ⎜
RBR−1 = ⎜ ⎝
λ1
⎞ λ2
..
.
⎟ ⎟ ⎠ λm
224
N. Nakauchi / Journal of Geometry and Physics 137 (2019) 217–227
where λ1 , . . . , λm ≥ 0. Since A = B −
RAR−1
1 (tr B) E, m
⎛ m 1 ∑ λj λ − 1 ⎜ m ⎜ j=1 ⎜ ⎜ ⎜ ⎜ = ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝
and hence RAR−1 = RBR−1 −
1 tr (RBR−1 ) E, m
we have
⎞ m
λ2 −
1 ∑ m
λj
j=1
..
. λm
⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟. ⎟ ⎟ ⎟ ⎟ m ⎟ 1 ∑ ⎠ λj − m
(14)
j=1
On the other hand, ˜ A is positive semi-definite and therefore so is A, since by (13), ˜ A is the inverse matrix of the symmetric, positive semi-definite matrix B. Then (14) leads us to get
λi −
m 1 ∑
m
λj ≥ 0
(15)
j=1
for any i. Let λi0 be a minimum of them, i.e., λi0 ≤ λi for any i. Then we have
λi0 =
1(
) ) (15) 1( λi0 + · · · + λi0 ≤ λ1 + · · · + λm ≤ λi0 . m m m
Then all the above inequalities are equalities, and hence λi = λi0 for any i. Therefore f ∗ h = λi0 g at P. Take the trace of 1 the both sides of this equality, and then we have ∥df ∥2 = λi0 m, i.e., λi0 = m ∥df ∥2 . Then f ∗ h = m1 ∥df ∥2 g at P. This is a contradiction to the assumption Tf ̸ = 0 at P. Thus we conclude Tf = 0 everywhere, i.e., f is a weakly conformal map. (2) ⇒ (1): If f is a weakly conformal map, then Tf = 0, therefore µf = 0 by the definition of µf . □ 5. Deformations of metrics and stress energy tensor In this section we prove that the stress energy tensor is a ‘‘gradient’’ of the functional Econf for deformations of metrics on the domain M. By moving the metric g, our functional Econf (f ) can be regarded as a functional of the metric g: Econf (f , g) =
∫
∥Tf ∥2g dvg , M
where the norm ∥Tf ∥ depends on the metric g and then we denote it by the notation ∥Tf ∥g . Theorem 2. Let gt be a 1-parameter family of metrics on M satisfying g0 = g. Then the functional Econf has the following first variation with respect to the metric g:
⏐ ⏐ Econf (f , gt ) ⏐⏐ dt
∫
d
( ) µf , w dvg ,
= 2 t =0
M
where
⏐ ⏐ ∂ ⏐⏐ ⏐ w = gt ⏐ = ∇ ∂ gt ⏐ . ∂t t =0 ∂ t t =0 def
Proof. We have d
⏐ ⏐
Econf (f , gt ) ⏐⏐ dt
(16) t =0
⏐ ∫ ⏐ d 2 = ∥Tf ∥g dvgt ⏐⏐ t dt t =0 ⏐ ) ⏐ ) ( ∫ (M ∫ ⏐ ⏐ ∂ ∂ 2 ⏐ = ∥Tf ∥g ⏐ dvgt + ∥Tf ∥2g dvgt ⏐⏐ t ∂t ∂t M M t =0 t =0
N. Nakauchi / Journal of Geometry and Physics 137 (2019) 217–227
225
Using (6), (7), (9) and (10), we define a tensor γ by
γ (X , Y ) =
m ∑
1
Tf (X , ej ) Tf (Y , ej ) +
m
j=1
=
m ∑
∥df ∥2 Tf (X , Y )
(17)
Tf (X , ej ) (f ∗ h)(Y , ej ) = h ξf (X ), df (Y )
(
)
j=1
=
m ∑
(f ∗ h)(X , ej ) Tf (Y , ej ) = h df (X ), ξf (Y ) .
(
)
j=1
Note that the stress energy tensor µf = 41 ∥Tf ∥2 g − γ . We use a local frame {ei } which is orthonormal with respect to the metric g = g0 and fix it. Then the tensor gt is represented by matrix (gt )ij with respect to the frame ei : (gt )ij = gt (ei , ej ) . Note that, when t = 0, gij = g(ei , ej ) = δij , where δij denotes the Kronecker’s delta. By Lemma 1(d), we have
∥Tf ∥2g = t
(
where (gt )
ij
m ∑
(gt )ik (gt )jℓ Tf (ei , ej ) (f ∗ h)(ek , eℓ ) ,
i, j, k, ℓ=1
)
(
)
is the inverse matrix of (gt )ij . Note
m ∑ ∂ ∂ (gt )ik = − (gt )ip (gt )pq (gt )qk ∂t ∂t p, q=1
since we differentiate both sides of the inverse relation m ∑
(gt )ip (gt )pq = δqi ,
p=1
where δqi denotes the Kronecker’s delta, i.e.,
δ = i q
{
1 0
if i = q if i ̸ = q
to have m m ∑ ∑ ∂ ∂ (gt )ip (gt )pq + (gt )ip (gt )pq = 0 ∂t ∂t p=1
p=1
and multiplying the inverse matrix (gt )qk , we have the above equality (18). Then we see
⏐ ⏐ ⏐ ik jℓ ∗ (gt ) (gt ) Tf (ei , ej ) (f h)(ek , eℓ ) ⏐⏐ ⏐ i, j, k, ℓ=1 t =0 ⏐ m ∑ ∂ ⏐ 2 (g )ik ⏐ g jℓ Tf (ei , ej ) (f ∗ h)(ek , eℓ ) ∂ t t ⏐t =0 i, j, k, ℓ=1 ⏐ ( ) m ∑ ⏐ qk ip ∂ ⏐ g −2 (g )pq g g jℓ Tf (ei , ej ) (f ∗ h)(ek , eℓ ) ∂ t t ⏐t =0 i, j, k, ℓ, p, q=1 ⏐ ⏐ m m ∑ ∑ ⏐ ∂ ∗ −2 Tf (ei , ej ) (f h)(ek , ej ) (g )ik ⏐ ∂ t t ⏐⏐ i, k=1 j=1 t =0 ⏐ m ∑ ⏐ ∂ −2 γ (ei , ek ) gt (ei , ek )⏐⏐ ∂t t =0 i, k=1 ⏐ ) ( ∂ ⏐⏐ −2 γ, g , ∂ t t ⏐t =0
⏐ ⏐ ∂ ∂ 2 ⏐ ∥Tf ∥g ⏐ = t ∂t ∂t t =0 = (18)
=
=
(17)
=
=
m ∑
(18)
226
N. Nakauchi / Journal of Geometry and Physics 137 (2019) 217–227
i.e.,
⏐ ⏐ ( ) ∂ ∥Tf ∥2g ⏐⏐ = −2 γ, w . t ∂t t =0
(19)
On the other hand we use here a local coordinate (x1 , . . . , xm ) to calculate represented by the following matrix (gt )ij
( (gt )ij = gt where
∂ (i ∂ xi
dvgt
∂ ∂ , ∂ xi ∂ xj
)
∂ dvgt ⏐t =0 ∂t
⏐
locally. The metric tensor gt is
,
= 1, 2, . . . , m) is the coordinate basis of the tangent space. Note √ ( √ ) = det (gt )ij dx1 · · · dxm and dvg = det(gij ) dx1 · · · dxm .
Lemma 4 provides us with
( )⏐ ⏐ m ∑ ( )⏐ ∂ (gt )ij ⏐⏐ ∂ ij det (gt )ij ⏐⏐ = g ⏐ ⏐ ∂t ∂t t =0 i,j=1
det(gij ) . t =0
Then we have
⏐ ⏐ √ ( ⏐ )⏐ ∂ ∂ ⏐ = dx1 · · · dxm dvgt ⏐ det (gt )ij ⏐⏐ ∂t ∂t t =0 t =0 ⏐ ) ( )⏐ ( 1 ∂ 1 = √ ( ) dx1 · · · dxm det (gt )ij ⏐⏐ 2 ∂ t t =0 det gij ⎛ ( ) ⏐⏐ ⎞ m ∑ ( ) ∂ (g ) 1 1 t ij ⏐ g ij = √ ( ) ⎝ ⏐ ⎠ det gij dx1 · · · dxm ⏐ 2 ∂t det gij i,j=1 t =0 ⏐ )√ ( ⏐ ( ) 1 ∂ ⏐ = trg g det gij dx1 · · · dxm 2 ∂ t t ⏐t =0 ) 1( = trg w dvg
(20)
2 ) 1( = g , w dvg , 2
where trg
(
∂ g ⏐ ∂ t t t =0
⏐
)
denotes the trace of the 2-tensor
⏐ ⏐ d Econf (f , gt ) ⏐⏐ dt t =0 ⏐ ∫ ⏐ d 2 = ∥Tf ∥g dvgt ⏐⏐ t dt t =0 ⏐ ∫ ∫ M ⏐ ∂ ∂ 2 ⏐ ∥Tf ∥2g = ∥Tf ∥g ⏐ dvg + dvgt t ∂ t ∂ t M t =0 ∫M ( ) 1 = − 2γ + ∥Tf ∥2 g , w dv 2 M ∫ (1 ) 2 ∥Tf ∥ g − γ , w dv =2 4 ∫M ( ) µ f , w dv . □ =2
∂ g ⏐ ∂ t t t =0
⏐
with respect to g. Therefore by (16), (19) and (20), we have
⏐ ⏐ ⏐ ⏐
t =0
M
In the rest of this section, we prove the following lemma which is used in the proof of Theorem 2.
(
)
Lemma 4. For any matrix-valued function A(t) = aij (t) with symmetry aij = aji , we have
⎛ ⎞ ( ) m ∑ ∂ det A ∂ A ∂ a ij ⎠ det A , = tr A−1 det A = ⎝ aij ∂t ∂t ∂t i,j=1
where (aij ) denotes the inverse matrix of A = (aij ).
(21)
N. Nakauchi / Journal of Geometry and Physics 137 (2019) 217–227
227
Proof. We first diagonalize the matrix A, i.e., there exists an orthogonal matrix P such that
⎛
λ1
PAP −1 = ⎝
⎜
⎞ ..
.
λm
⎟ ⎠.
Then we get
∂ ∂ det A = det (PAP −1 ) ∂t ∂t ) ∂ ( λ1 . . . λm = ∂ t ( ) 1 ∂λ1 1 ∂λm = + ··· + λ1 . . . λm λ1 ∂ t λm ∂ t ) ( ∂ (PAP −1 ) det A = tr (PAP −1 )−1 ∂t
(22)
Note that P depends on t, since A is a function of t. We have
) ∂ (PAP −1 ) ∂t { }) ( ∂ A −1 ∂ P −1 ∂ P AP −1 + P P + PA PA−1 P −1 ∂t ∂t ∂t ( ( )) ( ) ( ) ∂ P ∂ A ∂ P −1 −1 −1 −1 −1 −1 (PA P ) AP + tr PA P + tr P ∂t ∂t ∂t ) ) ( ) ( ) (( ∂ A −1 ∂ P −1 ∂ P −1 (PA−1 P −1 ) + tr PA−1 + tr P AP P ∂t ∂t ∂t ( ) ( ) ( ) ∂ P −1 ∂ P −1 ∂ A P + tr P + tr PA−1 P −1 ∂t ∂t ∂t ( ) ( ) −1 ∂ (PP ) ∂A + tr A−1 ∂t ∂t ( ) ∂ A A−1 . ∂t (
tr
= tr = tr = tr = tr = tr = tr
(PAP −1 )−1
(23)
Lemma 4 follows from (22) and (23), since
( tr
∂A A−1 ∂t
)
⎛ = ⎝
m ∑ i,j=1
⎞ ∂ a ij ⎠ aij ∂t
□
Acknowledgment This work was partially supported by the Grant-in-Aid for Scientific Research (C) No. 18K03280 at Japan Society for the Promotion of Science. References [1] P. Baird, J. Eells, A conservation law for harmonic maps, in: Geometry Symposium Utrecht 1980, in: Lecture Note in Math, vol. 894, Springer-Verlag, 1981, pp. 1–25. [2] J. Eells, L. Lemaire, A report on harmonic maps, Bull. Lond. Math. Soc. 10 (1978) 1–68. [3] J. Eells, L. Lemaire, Another report on harmonic maps, Bull. Lond. Math. Soc. 20 (1988) 385–524. [4] S. Kawai, N. Nakauchi, Weak conformality of stable stationary maps for a functional related to conformality, Differential Geom. Appl. 31 (2013) 151–165. [5] N. Nakauchi, A variational problem related to conformal maps, Osaka J. Math. 48 (2011) 719–741.