# Question about two ways to induce an inner product on $S^2V$

$\newcommand{\til}{\tilde}$

Let $(V,g)$ be an $n$-dimensional inner product space, and let $S^2V^*$ be the symmetric algebra. I am familiar with a natural way to endow $S^2V^*$ with an inner product via $g$.

In the paper The Riemannian Manifold of all Riemannian Metrics (by Gil-Medrano, Michor) a different way of endowing $S^2V^*$ with a product is described and I suspect it amounts to the same product I am familiar with.

Question: Are these inner products identical (maybe up to a scalar multiple)? (I will now describe the two approaches)

Approach 1:
We can think of $S^2V^*$ as the vector space of symmetric bilinear maps $V \times V \to \mathbb{R}$. As such, it is a subspace of $W:=\{T:V \times V \to \mathbb{R}| \, T \text{ is bilinear} \} \cong V^* \otimes V^*$ (The isomorphism is canonical). The inner ptoduct $g$ induces an inner product on $V^*$ which in turn induces a product on $W$. Now $S^2V^*$ inherits the inner product from $W$.

Approach 2 (from the paper):
Let $B \in S^2V^*=\{T:V \times V \to \mathbb{R}| \, T \text{ is bilinear and symmetric} \}$. $B$ induces a (linear) map $T_B:V \to V^*$ via $T_B(v)(\til v)=B(v,\til v)$. If $B$ is non-degenerate then $T_B$ is invertible. (In particular this is true for $T_g$).

Now, given $h,k \in S^2V^*$ , we define $\langle h,k \rangle_g =\operatorname{tr}(T_g^{-1}\circ T_h \circ T_g^{-1} \circ T_k)$

Note that $(T_g^{-1}\circ T_h \circ T_g^{-1} \circ T_k)$ is a linear map $V \to V$ so its trace is defined.

• Maybe link the paper. – IAmNoOne Apr 13 '16 at 16:53
• There's probably a more elegant approach to this question, but a low-brow approach would be to take an orthonormal basis $\beta = \{e_i \}_{i=1}^n$ for $(V,g)$, which generates a basis $\tilde{\beta} = \{e_i e_j \}_{i \leq j}$ for $S^2 V$. I think that in Approach 1, the basis $\tilde{\beta}$ is orthonormal, and I suspect that that is also true for Approach 2. One could check by writing down explicitly all the maps involved. Starting with a two-dimensional $V$ might make the computations a bit more concrete. – Phillip Andreae Apr 15 '16 at 2:43

The inner product on $V^*\otimes V^*$ is defined by extending bilinearly from simple tensors. Since $\mathrm{tr}(T_g^{-1}\circ T_h\circ T_g^{-1}\circ T_k)$ is also bilinear it suffices to check that they are equal when $h$ and $k$ are simple. Say $h=f\otimes f$ and $k=j\otimes j$. Let $\tilde{f}:\mathbb{R}\rightarrow V$ be the linear map that takes 1 to $f$, then $T_h=\tilde{f}\circ f$ and so $\mathrm{tr}(T_g^{-1}\circ T_h\circ T_g^{-1}\circ T_k)=$ $\mathrm{tr}(T_g^{-1}\circ \tilde{f}\circ f\circ T_g^{-1}\circ \tilde{j}\circ j)=$ $\mathrm{tr}(j\circ T_g^{-1}\circ \tilde{f}\circ f\circ T_g^{-1}\circ \tilde{j})=$ $(j\circ T_g^{-1}\circ f)(f\circ T_g^{-1}\circ j)=$ $\left\langle j,f\right\rangle\left\langle f,j\right\rangle$ which is the usual definition of inner product on $V^*\otimes V^*$.