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I get some big picture of tensor and tensor product by reading their Wikipedia articles, and several questions and answers posted before by others. But I cannot figure out how to show the following isomorphic equivalences:

  1. from Zach Conn:

    For finite-dimensional spaces V,W, the tensor product $V^* \otimes W$ is isomorphic to the space of homomorphisms $\textrm{Hom}(V, W)$. So in other words every linear map $V \to W$ has a tensor expansion, i.e., a representation as a tensor in $V^* \otimes W$.

    I wonder why "the tensor product $V^* \otimes W$ is isomorphic to the space of homomorphisms $\textrm{Hom}(V,W)$"?

  2. from Hans Lundmark:

    a bilinear map $B:V \times U \to \mathbb{R}$ can be canonically identified with an element of the space $V^* \otimes U^*$.

    How is a bilinear map $B:V \times U \to \mathbb{R}$ canonically identified with an element of the space $V^* \otimes U^*$?

  3. from Qiaochu Yuan:

    • A linear transformation $V \to V$,
    • An element of $V^* \otimes V$,
    • A linear map $V \otimes V^* \to \mathbb{R} $.

    The identification (of the first) with the second picture comes from the fact that dual distributes over tensor product (which again comes down to tensor contraction) and the fact that $V^{**} \cong V$. Alternately, again by tensor contraction, there is a natural bilinear map $V \times (V^* \otimes V) \to V$ which identifies an element of $V^* \otimes V$ with a linear transformation $V \to V$.

    How does that dual distributes over tensor product come from tensor contraction, and how does this lead to the identification?

    How is the natural bilinear map $V \times (V^* \otimes V) \to V$ defined?

Thanks and regards!

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Hint for 1. Try first to define a linear map from the tensor product to the Hom space. [Suggestions: define $R$, try to improve the formatting...] – Pierre-Yves Gaillard Aug 13 '11 at 7:38
@Tim: You should choose a definition of tensor product, otherwise it is not clear what the appropriate answer to each of your questions is. – Zhen Lin Aug 13 '11 at 7:43
Dear @Zhen. Thanks for the edit. Can't we assume that the tensor product is defined by the usual universal property? – Pierre-Yves Gaillard Aug 13 '11 at 7:56
@Zhen: The definition of tensor product I used is by a bilinear mapping with universal property. – Tim Aug 13 '11 at 12:43
up vote 10 down vote accepted

I'll attempt to sketch some of these maps below.

  1. $\DeclareMathOperator{\Hom}{Hom}$Let's try to define a linear map $\Phi\colon V^* \otimes W \to \Hom(V, W)$. We know that this is the same as giving a bilinear $V^* \times W \to \Hom(V, W)$. Now, given a functional $f\colon V \to \mathbf{R}$ and an element $w \in W$, I can produce a linear map $V \to W$ that sends $x \mapsto f(x)w$.

    To see that this is surjective [which is enough, since the dimensions match up], pick a basis $\{v_i\}$ of $V$ and a basis $\{w_i\}$ of $W$. Let $\{v_i^*\}$ be the dual basis to $\{v_i\}$. Then $$\Phi(v_i^* \otimes w_j)(v_k) = \delta_{ik}w_j.$$ So this sends $v_i$ to $w_j$, and sends all other $v_j$, $j \neq i$ to zero. These maps certainly span $\Hom(V, W)$.

  2. We want a bilinear map $V^* \times U^* \to \operatorname{Bilin}(V, U)$. If $f\colon V \to \mathbf{R}$ and $g\colon U \to \mathbf {R}$ are functionals, there is nothing more bilinear than multiplication: define $V \times U \to \mathbf{R}$ by $(x, y) \mapsto f(x)g(y)$.

  3. It's not clear to me what Qiaochu meant here, so I must be missing something. This actually follows from (2), since that part will show that giving an element of $V^* \otimes U^*$ is the same as giving a bilinear $V \times U \to \mathbf{R}$, and that's the same as an element of $(V \otimes U)^*$.

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+1!Very nice! About 1.: I would have avoided describing the inverse isomorphism. I would have used the fact the linear maps are sums of rank one linear maps to prove the surjectivity. I must be missing something ... – Pierre-Yves Gaillard Aug 13 '11 at 9:12
That is probably the nicest way to prove that it's an isomorphism, yes! But I remember the inverse having some significance. I'll look around. – Dylan Moreland Aug 13 '11 at 15:29

I'll note that there is a slightly different universal property the tensor product satisfies: for any three vector spaces $U, V, W$ over a common field $k$, there is an isomorphism $$\textrm{Hom}(U \otimes V, W) \cong \textrm{Hom}(U, \textrm{Hom}(V, W))$$ which is moreover natural in $U$ (contravariantly), $W$ (covariantly) and $V$ (extraordinarily). Taking $W = k$ yields the usual universal property in terms of bilinear maps. (A bilinear map $U \times V \to k$, after all, is essentially the same thing as a linear map $U \to \textrm{Hom}(V, k)$.)

The above adjunction property yields an alternative abstract-nonsense answer for the first question, if we take the isomorphism $(U \otimes V)^* \cong U^* \otimes V^*$ for granted. Recall that every finite-dimensional vector space is naturally isomorphic to its double dual. So, if $U$ and $V$ are finite-dimensional $k$-vector spaces, $$U^* \otimes V \cong \textrm{Hom}((U^* \otimes V)^*, k) \cong \textrm{Hom}(U^{**} \otimes V^*, k) \cong \textrm{Hom}(U^{**}, V^{**}) \cong \textrm{Hom}(U, V)$$ Obviously, this proof doesn't work when either $U$ or $V$ are infinite-dimensional. (In fact, the claim isn't even true.)

Dylan's answer adequately addresses the second question, so I'll skip that.

As for the third question, I too am not sure what Qiaochu meant. To me, tensor contraction is the linear map $U^* \otimes U \to k$ which is obtained from the bilinear map $U^* \times U \to k$ given by $(\alpha, v) \mapsto \alpha (v)$. If you think of $U^* \otimes U$ as being a space of matrices, this is just the trace map.

The natural bilinear map $\epsilon : V \times (V^* \otimes V) \to V$ Qiaochu is alluding to is the map given by $(v, \alpha \otimes v') \mapsto \alpha(v) v'$. In other words, it is the result of evaluating an element of $V^* \otimes V$, regarded as a linear map $V \to V$, at $v$. (But there is another natural bilinear map, namely the one given by $(v, \alpha \otimes v') \mapsto \alpha(v') v$. I'm quite sure this isn't the one he meant.) Another way of getting $\epsilon$ is by considering what happens to the map $V^* \otimes V \to \textrm{Hom}(V, V)$ from the first question under the isomorphism $$\textrm{Hom}((V^* \otimes V) \otimes V, V) \cong \textrm{Hom}(V^* \otimes V, \textrm{Hom}(V, V))$$

As for the connection with tensor contraction: note that there is a natural isomorphism $V \otimes (V^* \otimes V) \cong (V \otimes V^*) \otimes V$, and a natural isomorphism $V \otimes V^* \cong V^* \otimes V$; the map $\epsilon$ is essentially the composition of the trace map with these isomorphisms.

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Tensor contraction is the result of pairing together any factor of $V$ with any factor of $V^{\ast}$ in a tensor product of vector spaces under the evaluation map. The map I had in mind is a special case of the map $V \otimes (V^{\ast} \otimes W) \to W$ (so the first two factors are supposed to be paired). – Qiaochu Yuan Aug 13 '11 at 14:05

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