Does the Euclidean dot product (via the tensor product) induce the trace?

Question

Let $V = \mathbb{R}^n$, then the dot product $\cdot$ is a bilinear map $V \times V \to \mathbb{R}$. Thus, by the universal property of tensor products, a linear map $V \otimes V \to \mathbb{R}$ is induced.

If we identify $V \otimes V$ with $\operatorname{Hom}(V,V)=\operatorname{End}(V)=\operatorname{Mat}(n,\mathbb{R})=\mathfrak{gl}(n,\mathbb{R})=\mathscr{L}(\mathbb{R}^n)$, then I claim that this linear map induced by the tensor product is the trace. Is this correct?

Attempt: Let $e_1, \dots, e_n$ be the standard basis of $\mathbb{R}^n$.

We want the following diagram to commute: $$\begin{array}{rcl} V\times V & \overset{\cdot}{\to} & \mathbb{R} \\ \otimes & \searrow & \uparrow \\ & & V\otimes V \end{array}$$

Now $e_i \otimes e_j$ can be identified with the $n \times n$ matrix with all $0$'s except for a single $1$ in the $(ij)$th entry (I think). So if we want the diagram to commute, given $v, w \in V$, consider: $$v = v^i e_i, \quad w= w^ie_i, \quad v \cdot w = \sum_{i=1}^n v^i w^i, \quad v\otimes w = \sum_i \sum_j v^iw^j e_i \otimes e_j \,. $$ Thus our linear map $\operatorname{Mat}(n,\mathbb{R})$ has to be of the form: $$\sum_i \sum_j v^i w^j e_i \otimes e_j = (v^iw^j)_{ij} \mapsto \sum_i v^i w^i \,, $$ which is clearly just the sum of the diagonal entries and thus the trace.

Note: This is a follow-up to my previous question. In particular, I can't help but notice that both the Euclidean inner product and the evaluation map seem to induce the trace -- this can't be a coincidence. (Or maybe it can, but it doesn't seem like it.)

Also, the Lie algebra of the orthogonal group $\mathfrak{o}(n,\mathbb{R})$, which is the subspace of all matrices which preserve the Euclidean dot product $\otimes$, is a subspace of the special linear Lie algebra $\mathfrak{sl}(n,\mathbb{R})$, which is the kernel of the trace map, the linear map induced by the Euclidean dot product through the tensor product. This also seems like it can't be a coincidence.

Answer 1

By currying, we can make the connection between this question and your previous question a bit more obvious.

In particular: let $\phi_*:V \to V^*$ be defined so that $$ [\phi(v)](w) = v \cdot w $$ So, the map $$ V \times V \overset{\cdot}\to \Bbb R $$ can be broken down into the composition $$ V \times V \overset{ \phi_*\times \mathrm{id}}\to V^* \times V \overset{e}\to \Bbb R $$ Note that we need this map anyway to identify $V \otimes V$ with $Hom(V,V)$. In particular: the bilinear map $E: V^* \times V \to Hom(V,V)$ defined by $[E(f,v)](x) = vf(x)$ descends to a (natural) isomorphism $\tilde E: V^* \otimes V \to Hom(V,V)$. In order to get from $V \otimes V$ to $Hom(V,V)$, we need $$ V \times V \overset{ \phi_*\times \mathrm{id}}\to V^* \times V \overset{E}\to Hom(V,V) $$

So yes: your map is the trace, as long as you identify $V$ with $V^*$ in the manner described here. Note, however, that $\phi_*$ is not considered a "natural isomorphism". In particular, it "requires a choice" not intrinsic to the definition of a vector space.