# Prove there is a basis-independent isomorphism between $T^1_1(V)$ and the space of linear maps $V \rightarrow V$

Let $V$ be a finite-dimensional vector space. Prove there is a basis-independent isomorphism between $T^1_1(V)$ and the space of linear maps $V \rightarrow V$.

I want to show there is a bijective between elements in $T^1_1(V)$ and elements in $V \rightarrow V$, and them show there is a homomorphism. Mission failed.

Also, I read this question and do not understand: Dimension of Hom(U, V)

However, my question is more primitive - so regardless the complex discussion related to the two paper in the question and accepted answer, if the dimension of $\text{Hom}(U,V)$ just dim$U \times$ dim$V$?

Thanks!

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What is $T^1_1 (V)?$ – Ragib Zaman Jun 18 '13 at 16:15
Do I understand it correctly when I think that $T_1^1(V)$ stands for $V^*\otimes V$? In that case your question boils down to the well-known fact that $\text{Hom}(U,V)\cong U^*\otimes V$, where the correspondence can be given by $f\otimes v\mapsto (u\mapsto f(u)v)$. If not, I have to agree with @Ragib. – HSN Jun 18 '13 at 16:29
Does it have have to be basis-independent? I'd love to see that (not sarcasm at all: honest!) as I can't come up with something which doesn't appeal to basis... – DonAntonio Jun 18 '13 at 19:35
@DonAntonio: The explicit form given by HSN definitely does not make use of any basis. – celtschk Jun 18 '13 at 19:47
Thanks for that, @celtschk, yet I should have been more accurate in my comment: I meant how to prove it without resourcing say to basis and dual basis? For example, the proof that $\,V^{**}\cong V\,$ naturally doesn't use basis at all in the proof. I was kind of hoping something like that in this case... – DonAntonio Jun 18 '13 at 19:50

Let $a \in V$ and $\alpha \in V^*$. Let $A \in T^1_1$, so that $A(a, \alpha) \in K$, where $K$ is the base field of $V$.

There exists some derivative operator $\partial^{(\alpha)}$ that can be written in terms of a basis of vectors $e_1, e_2, \ldots$ and components of $\alpha$ labeled $\alpha_1, \alpha_2, \ldots$ as

$$\partial^{(\alpha)} = \sum_{i=1}^n e_i \frac{\partial}{\partial \alpha_i}$$

where $n$ is the dimension of $V$ (and $V^*$). While I have written this derivative operator in terms of a basis, the operator is itself basis independent; for instance, this operator could be defined using an integral (especially in the presence of a volume form or metric), or it could merely be shown to have the proper transformation laws of a vector.

Once this derivative is in place, we can construct a linear map $f: V \to V$ by

$$f(a) = \partial^{(\alpha)} A(a, \alpha)$$

Again, explicitly using a basis, this is

$$f(a) = \sum_{i=1}^n e_i A(a, e^i)$$

where $e^i$ are basis covectors.

Conversely, if there is a linear map $k: V \to V$, we can construct some $B \in T_1^1$ via the following. Use the identity

$$\partial^{(\alpha)} \alpha(a) = a$$

(This, again, can be proven in a basis, but the result is basis independent.) It's clear then that

$$\partial^{(\alpha)} (\alpha \circ k)(a) = k(a)$$

And thus, setting $B(a, \alpha) = (\alpha \circ k)(a)$ is consistent with the construction given earlier.

So far, we have proven the existence of these corresponding maps, but what about uniqueness? That comes from linearity. Suppose there were a tensor $C(a, \alpha)$ such that $\partial^{(\alpha)} C(a, \alpha) = f(a)$, then we would have

$$0 = \partial^{(\alpha)} A(a, \alpha) - \partial^{(\alpha)} C(a, \alpha)$$

Through linearity, we can rewrite this as

$$0 = \partial^{(\alpha)} (A-C)(a, \alpha)$$

This readily implies that $A = C$. Either they are both zero maps, or they are identical. A similar argument works the other way around.

To be honest, I do worry that this argument may not be basis-independent enough for you. I guess that will depend on how readily you can accept $\partial^{(\alpha)}$ being basis independent. This, I admit, is rather tricky; it's much much easier to construct it when you have a little more structure than just that of a vector space.

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