Tensor Problems

Lee Smooth Manifolds

Problem 1 (E12-6). (a) Let $\alpha$ be a covariant $k$-tensor on a finite-dimensional real vector space $V$. Show that $\mathrm{Sym}\alpha$ is the unique symmetric $k$-tensor satisfying

$$(\mathrm{Sym}\alpha )(v,\dots ,v)=\alpha (v,\dots ,v)$$

for all $v\in V$. (b) Show that the symmetric product is associative: for all symmetric tensors $\alpha ,\beta ,\gamma$,

$$(\alpha \beta )\gamma =\alpha (\beta \gamma ).$$

(c) Let ${\omega }^1,\dots ,{\omega }^k$ be covectors on a finite-dimensional vector space. Show that their symmetric product satisfies

$${\omega }^1\cdots {\omega }^k=\frac{1}{k!}\sum _{\sigma \in S_k}{\omega }^{\sigma (1)}\otimes \cdots \otimes {\omega }^{\sigma (k)}.$$

Proof) Clearly, the tensor $\mathrm{Sym}\alpha$ satisfies $\mathrm{Sym}\alpha (v,\dots ,v)=\alpha (v,\dots ,v)$ for all $v\in V$. The crux of this problem is showing uniqueness; by linearity, it suffices to show that if a symmetric covariant $k$-tensor $A$ satisfies $A(v,\dots ,v)=0$ for all $v\in V$, then $A=0$.

Fixing any arbitrary basis $(v_1,\dots ,v_n)$ of $V$; if we write $v=w^i{v}_i$, we have

$$A(v,\dots ,v)=A_{i_1,\dots ,i_k}{w}^{i_1}{w}^{i_2}\dots w^{i_k}$$

This expression collapses to zero for any $(w^1,\dots ,w^n)\in {\mathbb{R}}^n$. Instead of attacking the algebra head-on we opt for induction on $n$; the $n=1$ case is trivial as if $A_{1,1\dots ,1}({\omega }^1{)}^k=0$ for all ${\omega }^1\in \mathbb{R}$, clearly $A_{1,\dots ,1}=0$. Assuming $n-1$, fixing any one of the $w^i$s to zero and applying the $n-1$ case shows only coefficients of the form $A_{\sigma }$ where $\sigma$ is some permutation of $(1,2,\dots ,n)$ can be nonzero; these are all identical as $A$ is symmetric. Thus, $A$ is forced to be zero, and (a) is complete.

(b) is a trivial corollary of (a) as $(\alpha \beta )\gamma =\delta$ is the unique symmetric $k$-tensor satisfying

$$\delta (v,\dots ,v)=a\otimes \beta \otimes \gamma (v,\dots ,v)$$

for all $v\in V$, and the same can be said of the RHS. Note this is also satisfied by the triple symmetric product $\alpha \beta \gamma$, so we have $\alpha \beta \gamma =(\alpha \beta )\gamma =\alpha (\beta \gamma )$.

(c) follows from the observation that, for any $\sigma \in S_k$,

$${}^{\sigma }({\omega }^1\otimes \dots \otimes {\omega }^{\sigma (k)})(v_1,\dots ,v_k)=\prod _{i=1}^k{\omega }^i(v_{\sigma (i)})={\omega }^{{\sigma }^{-1}(1)}\otimes \dots \otimes {\omega }^{{\sigma }^{-1}(k)}(v_1,\dots ,v_k)$$

As the $S_k$ is a group under the composition of permutations, the expression on the right is uniformly distributed across $S_k$ and is equivalent to the symmetric product ${\omega }^1\dots {\omega }^k$.

Remark: It is a special property of symmetric tensors that they are entirely determined by the values $A(v,\dots ,v)$. For instance, there are infinitely many alternating tensors, but for any $k\ge 2$ the expression $A(v,\dots ,v)$ always evaluates to zero.

Problem 2 (E12-8). Let $M$ be a smooth $n$-manifold, and let $A$ be a smooth covariant $k$-tensor field on $M$. If $\left(U,\left(x^i\right)\right)$ and $\left(\tilde{U},\left({\tilde{x}}^j\right)\right)$ are overlapping smooth charts on $M$, we can write

$$A=A_{i_1\dots i_k}d{x}^{i_1}\otimes \cdots \otimes d{x}^{i_k}={\tilde{A}}_{j_1\dots j_k}d{\tilde{x}}^{j_1}\otimes \cdots \otimes d{\tilde{x}}^{j_k}.$$

Compute a transformation law expressing the component functions $A_{i_1\dots i_k}$ in terms of ${\tilde{A}}_{j_1\dots j_k}$.

Proof) Fix any $p\in U\cap \tilde{U}$. The covariant $k$-tensor $A_p$ is a multilinear functional on $(T_{p}M{)}^k$, and thus descends to a linear map from $(T_{p}M)\otimes \dots \otimes (T_{p}M)\to \mathbb{R}$. “Decomposing” tuples of tangent vectors is a well-defined operation in this tensor product space.

It suffices to represent ${\xi }^{j_1,\dots ,j_k}=d{\tilde{x}}_p^{j_1}\otimes \dots \otimes d{\tilde{x}}_p^{j_k}$ as a linear combination of ${\eta }^{i_1,\dots ,i_k}=d{x}_p^{i_1}\otimes \dots \otimes d{x}_p^{i_k}$, and thus it suffices to represent the vector $w^{i_1,\dots ,i_k}=(\partial /\partial x_p^i)$ as a linear combination of $v^{j_1,\dots ,j_k}=(\partial /\partial {\tilde{x}}_p^j)$. Let $\Phi$ be the identity diffeomorphism $U\cup \tilde{U}\to U\cup \tilde{U}$ and $D\Phi$ be its Jacobian at $p$ with the domain in $U$ coordinates and the codomain in $\tilde{U}$ coordinates. Then,

$$w^{i_1,\dots ,i_k}={\left(\frac{\partial }{\partial x_p^{i_l}}\right)}_{l=1}^k={\left(D{\Phi }_{m,i_l}\frac{\partial }{\partial {\tilde{x}}_p^m}\right)}_{l=1}^k=\sum _{m\in \{1,2,\dots ,n{\}}^k}\left(\prod _{l=1}^{k}D{\Phi }_{m_l,i_l}\right)\left(\underset{l=1}{\overset{k}{⨂}}\frac{\partial }{\partial {\tilde{x}}_p^{m_l}}\right)$$

Therefore,

$${\xi }^{j_1,\dots ,j_k}{w}^{i_1,\dots ,i_k}=\prod _{l=1}^{k}D{\Phi }_{j_l,i_l}\Rightarrow {\xi }^{j_1,\dots ,j_k}=\sum _{i\in \{1,\dots ,n{\}}^k}\left(\prod _{l=1}^{k}D{\Phi }_{j_l,i_l}\right){\eta }^i$$

and

$$A_{i_1,\dots ,i_k}=\sum _{j\in \{1,\dots ,n{\}}^k}{\tilde{A}}_j\left(\prod _{l=1}^{k}D{\Phi }_{j_l,i_l}\right)$$

Remark: Note that there is a canonical isomorphism between multilinear functionals on $(T_{p}M{)}^k$ and linear functionals on the $k$-abstract tensor product $T_{p}M\otimes \dots \otimes T_{p}M$, and so the covariant $k$-tensor space on $p$ can be identified with the space of linear functionals on $T_{p}M\otimes \dots \otimes T_{p}M$; we use this identification multiple times throughout this computation.

Remark 2: This transformation law can be easily generalized to mixed tensor fields by identifying the tensor space $T^k(V)$ with $L(V^{\ast },\dots ,V^{\ast };\mathbb{R})$ and applying the inverse Jacobian to covectors.

Problem 3 (E12-12). Let $M$ be a smooth manifold and $V\in \mathfrak{X}(M)$. Show that the Lie derivative operators on covariant tensor fields, ${\mathcal{L}}_V:{\mathcal{T}}^k(M)\to {\mathcal{T}}^k(M)$ for $k\ge 0$, are uniquely characterized by the following properties: (a) ${\mathcal{L}}_V$ is linear over $\mathbb{R}$. (b) ${\mathcal{L}}_{V}f=Vf$ for $f\in {\mathcal{T}}^0(M)=C^{\infty }(M)$. (c) ${\mathcal{L}}_V(A\otimes B)={\mathcal{L}}_{V}A\otimes B+A\otimes {\mathcal{L}}_{V}B$ for $A\in {\mathcal{T}}^k(M)$, $B\in {\mathcal{T}}^l(M)$. (d) ${\mathcal{L}}_V(\omega (X))=({\mathcal{L}}_V\omega )(X)+\omega ([V,X])$ for $\omega \in {\mathcal{T}}^1(M)$, $X\in \mathfrak{X}(M)$.

Proof) The Lie derivative operator trivially satisfies all four properties. Assume some other operator $\xi$ satisfies all four; by (b), ${\mathcal{L}}_{V}f=Vf=\xi f$ for all $f\in {\mathcal{T}}^0(M)$. Therefore, $({\mathcal{L}}_V\omega )(X)+\omega ([V,X])={\mathcal{L}}_V(\omega (X))=\xi (\omega (X))=(\xi \omega )(X)+\omega ([V,X])\Rightarrow \xi \omega ={\mathcal{L}}_V\omega$. Thus, $\xi \omega ={\mathcal{L}}_V\omega$ for all $\omega \in {\mathcal{T}}^1(M)$. For any $k\ge 2$ and $k$-tensor $v_1\otimes \dots \otimes v_k$,

$$\xi (v_1\otimes \dots \otimes v_k)=\xi (v_1)\otimes \dots \otimes v_k+\dots +v_1\otimes \dots \otimes \xi (v_k)$$ $$={\mathcal{L}}_V(v_1)\otimes \dots \otimes v_k+\dots +v_1\otimes \dots \otimes {\mathcal{L}}_V(v_k)={\mathcal{L}}_V(v_1\otimes \dots \otimes v_k)$$

Thus, $\xi$ and ${\mathcal{L}}_V$ are linear operators (by (a)) ${\mathcal{T}}^k(M)\to {\mathcal{T}}^k(M)$ that agree on $k$-tensors of the form $v_1\otimes \dots \otimes v_k$. As these span ${\mathcal{T}}^k(M)$, they are forced to be identical.

Remark: Similar to the axiomatic definition of determinants, the Lie derivative of tensor fields can be defined in a flow-independent way using these properties. This allows it to be generalized to vector fields on manifolds with boundary on which a flow does not necessarily exist.