Let $\pi=E\rightarrow B$ be a smooth vector bundle, a fibre metric $g_E$ is a section of $\mathcal{Sym}^2(E)$ such that $g_E(p)$ defines an inner product on $\pi^{-1}(p)$ for every $p\in B$ which varies smoothly, the latter just means that for any two sections $X,Y$ of $E$ we have that $g_E( X,Y )$ is $C^\infty$. Now I need to prove the following statement

Suppose $(M,g)$ is a Riemannian manifold, then for any tensor bundle $\mathcal{J}_l^k(M)$ there is a unique fibre metric $\langle \cdot,\cdot\rangle $ such that if $\{e_i\}$ is an orthonormal basis for $T_pM$ with dual basis $\{e^{i} \}$, then the collection of tensors $$e_{i_1}\otimes\cdots\otimes e_{i_l}\otimes e^{j_1}\otimes\cdots \otimes e^{j_k} \quad \quad \text{(1)}$$ form an orthonormal basis for $\mathcal{J}_l^k(T_pM)$.

Here are some of my thoughts on this

  • The set defined on (1) is clearly a basis
  • This must have something to do with the sharp and flat maps, which are defined as $X^\flat Y=g(X,Y)$ and $A\mapsto A^\sharp$ such that $AY=g(A^\sharp,Y)$ for $X,Y$ vector fields and $A$ covariant tensor field.
  • In local coordinates these maps look like $$X^\flat=\sum_j X_j dx^j,\quad X_j=\sum_i g_{ij}X^i$$ $$A^\sharp=\sum_j A^j \partial_j, \quad A^j=\sum_i g^{ij}A_i $$ which is the reason we say $g$ is used to ''low'' and ''rise'' indices, this clearly be used. For example, let $T,S\in \mathcal{J}^k(M)$ defined in some neighborhood of $p\in M$, given a basis as states above, we can write (using Einstein's convention, otherwise the summation indices will be simply horrifying) $$T=T^{i_1\ldots i_l}_{j_1,\ldots,j_k}e_{i_1}\otimes\cdots\otimes e_{i_l}\otimes e^{j_1}\otimes\cdots \otimes e^{j_k}$$

$$S=S^{i_1\ldots i_l}_{j_1,\ldots,j_k}e_{i_1}\otimes\cdots\otimes e_{i_l}\otimes e^{j_1}\otimes\cdots \otimes e^{j_k}$$ Our natural choice would now be focus on the components in such a way that we multiply them if their indices correspond to each other, in some kind of $\delta^{i_1\ldots i_l}_{j_1,\ldots,j_k}$ way, but I do not know how to write this down. I believe the uniqueness and smoothness will follow trivially after the precise definition of this inner product. Any help is appreciated.

Thanks in advance!


1 Answer 1


First lets talk about how this works on vector spaces and the generalization to vector bundles (at least locally) isn't too difficult. Let $V$ and $W$ be vector spaces with inner products $\langle \cdot , \cdot \rangle_V $ and $\langle \cdot , \cdot \rangle_W $ respectively. There is a natural inner product on $V \otimes W$ mainly

$$\langle v_1 \otimes w_1 , v_2 \otimes w_2\rangle = \langle v_1, w_1 \rangle_V \langle v_2, w_2 \rangle_W $$

Clearly if a basis $\{v_i\}$ is orthonormal wrt $\langle \cdot , \cdot \rangle_V $, and $\{w_j\}$ is orthonormal wrt $\langle \cdot , \cdot \rangle_W $ then the basis $\{ v_i \otimes w_j\}$ is orthonormal wrt $\langle \cdot ,\cdot \rangle$.

There is also a naturally induced inner product on the dual space $g^\sharp$ such that $\forall \alpha, \beta \in V^*$

$$g^\sharp ( \alpha, \beta) := g( \alpha^\sharp , \beta^\sharp) $$

When given an inner product $g$ on $V$. Now one can also show that if $\{v_i\}$ is orthonormal wrt $g$, the dual basis $\{v^i\}$ are orthonormal wrt $g^\sharp$.

Using both these ideas we can induce a metric on $T^k_l (V):=\overbrace{V\otimes \cdots \otimes V}^l\otimes \overbrace{V^\ast\otimes \cdots \otimes V^\ast}^k$ which guarantees that the natural basis on the space is orthonormal. Now the only remaining tasks would be to show that given a fixed orthonormal basis for $(V,g)$ the induced metric on $T^k_l(V)$ is the unique one that guarantees the induced basis on $T^k_l(V)$ is orthonormal and smoothness. Hope this helps.


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