Tensor characterization

Question

I want to show that $T^{1}_{1}(V)$ is isomorphic to $End(V)$. I know how to produce a linear map from $h:End(V)$ to $T^{1}_{1}(V)$, i.e. send $f \in End(V)$ to $hf(w,v) = w(f(v))$, but how to write down its inverse explicitly since I want to calculate the trace of its inverse?

Answer 1

For any vector space $X$, write "$X^T$" for its dual.

Then we define $g:T^1_1(V)\to \mathrm{Hom}(V,V^{TT})$ by $((gb)v)w = b(w,v)$.

When $V$ is finite dimensional we have $V^{TT}\cong V$, and then $\mathrm{Hom}(V,V^{TT})\cong\mathrm{End}(V)$ and $g$ gives the inverse to $h$.

Answer 2

The trick here is that this fact holds only for finite-dimensional vector spaces $V$, so, in order to establish it, we are forced to deal with some kind of a "proof of dimensionality", e.g. a basis.

Notice, that elements of $T^1_1(V)$ are finite sums of "simple" tensors, that is, objects of the form $v \otimes \phi$ for some vector $v \in V$ and a "covector" $\phi \colon V \to k$ (a linear map) where $k$ is the base field, e.g. $\mathbb{R}$ or $\mathbb{C}$.

Suppose, that $\{ e_i \}_{i=1,\dots,n}$ is a basis in $V$, where $n = \dim V$.

We have the dual basis $\{ \widehat{e}_j \}$ in $T_1 := V^* := Hom(V,k)$ and the elements in the dual basis are defined on $v = \sum_i v^i e_i$ by $ \widehat{e}_j (v) = v^j$.

Now, any element $t \in T^1_1(V)$ is uniquely represented as $$ t = \sum_{1 \leq j,k \leq n} a_{jk} e_j \otimes \widehat{e}_k \tag{*} $$

(Think of a matrix $A = (a_{ij})$ of the endomorphism that we a looking for!)

It is clear, that for $t$ as in $(*)$ we can assign an endomorphism $$ v = \sum_{i} v^i e_i \mapsto \sum_{1 \leq j,k \leq n} a_{jk} e_j \otimes \widehat{e}_k (v) = \sum_{1 \leq j,k \leq n} a_{jk} v^k e_j \tag{**} $$

It remains to check that $(*) \to (**)$ indeed yields the inverse to your mapping.

I leave the verifications and final conclusions up to you. If I have time later, I will try to update this answer, and. in particular, improve the notation.