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I'm trying to learn tensor products by myself, and I'm using primarily the Linear Algebra and Geometry book, by Kostrikin and Manin. I need help with the following question:

Let $V$ and $W$ be $\mathbb{K}$-vector spaces. Show that there is a unique linear transformation $\Gamma: V \otimes W \rightarrow \mathcal{L}(V^*,W)$ that satisfies $$\Gamma(v\otimes w)(f) = f(v)w$$ for any $v \in V, w \in W$ and $f\in V^*$.

My first (and only) idea was to define an application $t$ from $V\times W$ to $V \otimes W$ that associates the pair $(v,w)$ to $v\otimes w$, and use (somehow) that $t$ is universal. However, I'm very much stuck.

Any help would be deeply appreciated.

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    $\begingroup$ Assuming $\mathcal{L}(V^*,W)$ means linear maps $V^*\to W$ then define $\Gamma:V\times W\rightarrowtail \mathcal{L}(V^*,W)$ by $\Gamma(v,w):f\mapsto f(v)w$. Prove $\Gamma$ is biadditive (distributes) and apply universal mapping property of the tensor product. $\endgroup$ – Algeboy Jul 25 at 19:50
  • $\begingroup$ Thank you, @Algeboy. I've managed to answer the question. If no one post an answer, I will post mine latter on, just to close the question. $\endgroup$ – Bruno Tassone Jul 25 at 20:17
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Recall that every pair of $\Bbb K$-vector spaces $V$ and $W$ defines a map

enter image description here

where $\Phi$ is the $\Bbb K$-bilinear map defined by $\Phi(v, w)=v\otimes w$. The map $\Phi$ is characterized by the following universal property.

Universal Property of $\Phi$. Consider a diagram of $\Bbb K$-vector spaces

enter image description here

where $\phi$ is $\Bbb K$-bilinear. Then there exists a unique $\Bbb K$-linear map $\Gamma$ making

enter image description here

commute.

Next, to address your example, take $X=\mathcal L(V^\ast, W)$. The space $V^\ast$ consists of all linear maps $V\to \Bbb K$ and the space $\mathcal L(V^\ast, W)$ consists of all linear maps $V^\ast \to W$.

Now, let $\phi:V\times W\to\mathcal L(V^\ast, W)$ be defined by $\phi(v, w)(f)=f(v)\cdot w$. The formulas \begin{align*} \phi&(\lambda_1\cdot v_1+\lambda_2\cdot v_2, w)(f)\\ &= f(\lambda_1\cdot v_1+\lambda_2\cdot v_2)\cdot w & \phi&(v, \lambda_1\cdot w_1+\lambda_2\cdot w_2)(f)\\ &= f(v)\cdot(\lambda_1\cdot w_1+\lambda_2\cdot w_2) \\ &= \{\lambda_1\cdot f(v_1)+\lambda_2\cdot f(v_2)\}\cdot w & &= \lambda_1\cdot f(v)\cdot w_1+\lambda_2\cdot f(v)\cdot w_2 \\ &= \lambda_1\cdot f(v_1)\cdot w+\lambda_2\cdot f(v_2)\cdot w & &= \lambda_1\cdot\phi(v, w_1)(f)+\lambda_2\cdot\phi(v, w_2)(f) \\ &= \lambda_1\cdot\phi(v_1, w)(f)+\lambda_2\cdot\phi(v_2, w)(f) \end{align*} demonstrate that $\phi$ is $\Bbb K$-bilinear.

Now, we have a diagram

enter image description here

where $\phi$ is bilinar. What does the universal property of $\Phi$ say?

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