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For $V$ a finite dimensional vector space over a field $\mathbb{K}$, I have encountered the claim that $$ \dim(\mathrm{Hom}(V,V)) = \dim(\mathrm{Hom}(V \times V, \mathbb{K})) $$

where $\mathrm{Hom}(V,V)$ denote the vector spaces, respectively, of all linear maps from $V$ to $V$ and all bilinear maps from $V\times V$ to the ground field $\mathbb{K}$. I'm sure I'm overlooking something elementary, but I don't see this.

There is a theorem that, in general, for any finite-dimensional vector spaces $V$ and $W$ that $$ \dim(\mathrm{Hom}(V,W)) = \dim(V)\dim(W) $$

But, $\dim(V \times W) = \dim(V) + \dim(W)$ and therefore $$ \dim(\mathrm{Hom}(V \times V, \mathbb{K})) = (\dim(V) + \dim(V))\cdot \dim(K) = 2\dim(V)\cdot 1 $$ which is obviously not equal to $\dim(\mathrm{Hom}(V,V)) = \dim(V)\cdot\dim(V)$

Where is my mistake?

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A bilinear map from $V \times V$ to $\mathbb{K}$ is not the same thing as a linear map from $V \times V$ to $\mathbb{K}$. You have computed the dimension of the latter. –  Chris Eagle May 30 '12 at 16:01
    
Ok, so I guess I need to know how to calculate the dimension of the vector space of bilinear (n-linear?) maps to the ground field. Can you suggest a reference that discusses this? –  AFX May 30 '12 at 16:09
    
@AFX Here's a googlebooks link that looks very helpful. –  rschwieb May 30 '12 at 16:12
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@AFX: You really should stop using $\mathrm{Hom}$ for both linear maps and bilinear maps. That's what got you into trouble in the first place. Assuming the first $\mathrm{Hom}$ refers to bilinear, what I'm saying is that those two spaces are not only isomorphic, but even naturally isomorphic. –  Chris Eagle May 30 '12 at 16:21
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@AFX I seen Bilin(-,-) and Bihom(-,-) both used, and your suggestion is OK as long as you alert the reader to what it means. –  rschwieb May 30 '12 at 16:52

2 Answers 2

up vote 3 down vote accepted

I would not write $\operatorname{Hom}(V \times V, \mathbb K)$ for the space of bilinear maps, since there is nothing to distinguish this from your old notation for the space of linear maps. I've seen $L^2(V, V; \mathbb K)$ used, but $\operatorname{Bilin}(V, V; \mathbb K)$ has the advantage of being obvious. In any case, it never hurts to specify your notation.

Now to calculate the dimension. Let $\{e_1, \ldots, e_n\}$ be a basis for $V$, and let $\{f_i\}$ be the corresponding dual basis. Then I claim that the set of $n^2$ bilinear maps \[ F_{ij}(x, y) = f_i(x)f_j(y) \qquad i, j = 1, \ldots, n \] is a basis for $L^2(V, V; \mathbb K)$. To remember this fact, it might help to recall how bilinear forms correspond to matrices after one has chosen a basis.

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Chris Eagle has already comment-answered the main problem, but I'm also going to add that the set $Bilin_{\mathbb{K}}(V\times V,\mathbb{K})$ is another way of writing $V\otimes_{\mathbb{K}}V$, and this latter guy has dimension $dim(V)^2$. This is the interpretation of tensors as multilinear functionals.

EDIT: As Chris was nice enough to remind me, $Bilin_{\mathbb{K}}(V\times V,\mathbb{K})$ is actually naturally identified with the dual module $(V\otimes_{\mathbb{K}}V)^\ast$ rather than just $V\otimes_{\mathbb{K}}V$. But since finite dimensional vector spaces are isomorphic to their duals, the statement about dimensions is still OK.

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Surely the space of bilinear maps is the dual of the tensor product? They have the same dimension, of course, but they are not the same thing. –  Chris Eagle May 30 '12 at 16:12
    
@ChrisEagle Hmm you might be right! I had thought the tensor product recorded bilinear maps, but maybe I left soemthing out. I remember some jiggering around when V had an inner product and I might have mixed it up. Double checking and revising after lunch. –  rschwieb May 30 '12 at 16:13
    
You still see this first definition of $V \otimes V$ in most differential geometry books, I'm afraid. –  Dylan Moreland May 30 '12 at 16:52
    
Chris is correct. The tensor product is the universal thing through which bilinear maps factor rather than the thing which describes bilinear maps directly. Some differential geometry books do not use this convention; I think it is out of a desire to express a tensor product-like construction as an explicit space of maps rather than as an abstract universal thing, but ultimately it seems to me that it just leads to more confusion. –  Qiaochu Yuan May 30 '12 at 21:12
    
@QiaochuYuan Let me know if the edit in place is not enough to alert people that this was already corrected. –  rschwieb May 30 '12 at 21:21

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