# Uniqueness of fibre metric in Riemannian Manifold

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.

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.