I am trying to prove Lemma 3.1 in Lee's book "Riemannian manifolds", which is left as an exercise. The Lemma goes as follows:

Let $(M,g)$ be a Riemannian manifold. Show that there exists a unique fiber metric on the tensor bundle $T_l^k M$ such that if $(E_1,...,E_n)$ is an orthonormal basis for $T_p M$ and $(\phi^1,..., \phi^n)$ is the dual basis, then the tensors of the form:

$E_{j_1} \otimes E_{j_2} \otimes .. \otimes E_{j_l} \otimes \phi^{i_1}\otimes...\otimes \phi^{i_k}$

form an orthonormal basis of $T_l^k(T_p M)$.

As a hint, the inner product on local coordinates is given by:

$\langle F, G \rangle = g^{i_1 r_1} \cdot \cdot \cdot g^{i_k r_k} g_{j_1 s_1}\cdot \cdot \cdot g_{j_l s_l} \: F^{j_1...j_l}_{i_1...i_k} \:G^{s_1...s_l}_{r_1...r_k}$.

where $F^{j_1...j_l}_{i_1...i_k}$ and $G^{s_1...s_l}_{r_1...r_k}$ are the components w.r.t the basis $E_{j_1} \otimes E_{j_2} \otimes .. \otimes E_{j_l} \otimes \phi^{i_1}\otimes...\otimes \phi^{i_k}$ of $F$ and $G$.
And $g_{ij}$ are the matrix entries of the metric tensor, $g^{ij}$ the ones of the inverse.

My ideas:

It is clear that this product is bilinear. Since $g_{ij}$ is positive definite, and $g^{ij}$ as its inverse is too, so is $\langle \cdot , \cdot \rangle$. I do not understand why this basis is orthonormal. But it is clear to me why the metric is uniquely determined by an orthonormal basis.


1 Answer 1


Interesting, in the newest edition of Lee's "Riemannian Manifolds" this is Proposition 2.40, and perhaps it gives a nicer hint. Let's get a few preliminaries down.

On a Riemannian manifold $M$ we have a metric $g$. The inner product is given as

$$ \langle X,Y\rangle=g_{ij}X^iY^j $$

for $TM$ with respect to the coordinate frame.

Similarly, by considering the canonical bundle isomorphism $\tilde{g}:TM\to T^*M$ given by the Riemannian metric, we may construct a bundle metric on $T^*M$ as follows. For any $\eta,\xi\in T^*M$, there exists unique $X,Y\in TM$ such that

$$ \eta=g(X,\cdot),\;\;\xi=g(Y,\cdot), $$

and we then define

$$ \langle\eta,\xi\rangle=\langle X,Y\rangle. $$

We also have that

$$ \langle\eta,\xi\rangle=g^{ij}\eta_i\xi_j $$

with respect to the coordinate frame.

Given a local orthonormal frame $(E_j)$ for $TM$, it may be checked that the dual frame $(\phi^i)$ is actually a local orthonormal frame for $T^*M$.

Now consider the tensor bundle $T^k_\ell(M)$ and define the bundle metric as follows. For any (pure) $(\ell,k)$-tensors

\begin{align*} X_1\otimes\cdots\otimes X_\ell\otimes\eta^1\otimes\cdots\otimes\eta^k \\ Y_1\otimes\cdots\otimes Y_\ell\otimes\xi^1\otimes\cdots\otimes\xi^k \end{align*}

in $T^k_\ell(M)$ we define

$$ \langle X_1\otimes\cdots\otimes X_\ell\otimes\eta^1\otimes\cdots\otimes\eta^k,Y_1\otimes\cdots\otimes Y_\ell\otimes\xi^1\otimes\cdots\otimes\xi^k\rangle=\langle X_1,Y_1\rangle\cdots\langle \eta^k,\xi^k\rangle. $$

We then extend this bilinearly and this defines a bundle metric on $T^k_\ell(M)$. Note, this is in the statement of the proposition, and is what I consider the "hint".

With this definition, we may check that it satisfies your definition in coordinates, hence it is equivalent. Finally, by the statement above about the local orthonormal frames, and the "linearity over the tensor product", you should be able to check that the tensor products of basis elements determines a local orthonormal frame for the tensor bundle.


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