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This is exercise 12-4 in the book Introduction to Smooth Manifolds by John M. Lee:

Let $V_1,\dots,V_k$ and $W$ be finite-dimensional real vector spaces. Then $$V_1^* \otimes \dots \otimes V_k^* \otimes W \cong \mathrm{L}(V_1,\dots,V_k;W)$$ canonically, where $\mathrm{L}(V_1,\dots,V_k;W)$ denotes the set of multilinear mappings from $V_1\times \dots \times V_k$ to $W$.

It is clear, that we use the universal property of the tensor product space, so we construct a mapping $$\Phi : \begin{cases} V_1^* \times \dots \times V_k^* \times W \to \mathrm{L}(V_1,\dots,V_k;W)\\ (\varphi_1,\dots,\varphi_k,w) \mapsto \left((v_1,\dots,v_k) \mapsto \varphi_1(v_1) \cdots \varphi_k(v_k)w\right) \end{cases}$$

Now $\Phi$ is multilinear and thus we get a linear mapping $$\widetilde{\Phi} : V_1^* \times \dots \times V_k^* \times W \to \mathrm{L}(V_1,\dots,V_k;W)$$ such that $\Phi = \widetilde{\Phi} \circ \otimes$. It is easily checked that $\ker\widetilde{\Phi} = \{0\}$ and thus $\widetilde{\Phi}$ is injective. How do we show, that $\widetilde{\Phi}$ is surjective not knowing what the dimension of $\mathrm{L}(V_1,\dots,V_k;W)$ is? I want to use this exercise to determine the dimension of $\mathrm{L}(V_1,\dots,V_k;W)$.

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  • $\begingroup$ You could try make it for $k = 1$ and make canonical isomorphism. Just select any vector and consider pair $(f(x),x))$, where $x$ our vector. After it you could figure that $f(x)$ is some scalar from field. And after this you should prove injective and surjective of this map. $\endgroup$ – openspace Jul 2 '17 at 10:47
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Suppose you have $f\in\operatorname{L}(V_1,\dots,V_k;W)$. By the universal property, you get a linear map $\hat{f}\colon V_1\otimes\dots \otimes V_k\to W$ and conversely, so what you really have to show is

  1. $L(V,W)\cong V^*\otimes W$
  2. $(V_1\otimes\dots \otimes V_k)^*\cong V_1^*\otimes\dots \otimes V_k^*$

Note that 2 is a special case of the theorem, but with $W=\mathbb{R}$ (in general, the base field), where you just need injectivity and the fact that the dimension of the two spaces is clearly the same.

For 1 it's again proving injectivity and equality of dimensions.

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  • $\begingroup$ Thanks. Additionally, I have also to show that if $V_1 \cong V_2$, then $V_1 \otimes W \cong V_2 \otimes W$ for bringing together your $2$ isomorphisms, right? Also I do not quite see yet the "conversly" part of your first sentence. $\endgroup$ – TheGeekGreek Jul 2 '17 at 11:54
  • $\begingroup$ @TheGeekGreek “Conversely” refers to the fact that from a linear map $g\colon V_1\otimes\dots \otimes V_k\to W$ you get a unique element in $\operatorname{L}(V_1,\dots,V_k;W)$. The first statement in your comment is obvious, isn't it? $\endgroup$ – egreg Jul 2 '17 at 12:04
  • $\begingroup$ Yes, my first statement is obvious, I just wanted to be sure. From what exactly does this follow? I mean given $g : V_1 \otimes \dots \otimes V_k \to W$ linear, then there exists a unique $\widehat{g} : V_1 \times \dots \times V_k \to W$ multilinear. $\endgroup$ – TheGeekGreek Jul 2 '17 at 12:09
  • $\begingroup$ @TheGeekGreek $\hat{g}(v_1,\dots,v_k)=g(v_1\otimes\dots\otimes v_k)$. $\endgroup$ – egreg Jul 2 '17 at 12:11
  • $\begingroup$ Thank you! Sometimes the easiest things are the hardest. So in conclusion we do have $$\mathrm{L}(V_1 \otimes \dots \otimes V_k;W) \cong \mathrm{L}(V_1 \times \dots \times V_k ; W)$$ right? $\endgroup$ – TheGeekGreek Jul 2 '17 at 12:18

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