# The obvious Tensorhom Identification!

Let's say you have a vector space $V$ and a vector space $W$ over the same field $k$. Then, according to a huge pile of books on representation theory I read, there is an obvious isomorphism $V^\ast\otimes_k W \cong \mathrm{Hom}_k(V,W)$, where $V^\ast=\mathrm{Hom}_k(V,k)$. I concur, it's not hard to write down, but then why don't they just write it down? It is given by $\begin{array}{rcl} \phi: V^\ast\otimes_k W &\longrightarrow& \mathrm{Hom}_k(V,W) \\ f\otimes w &\longmapsto& (v\mapsto f(v)w), \end{array}$ when we assume $W$ to be finite-dimensional. To construct an inverse, simply choose a basis $B$ for $W$ and consider the map that sends $g\in\mathrm{Hom}_k(V,W)$ to $\psi(g) := \sum_{b\in B} (b^\ast\circ g)\otimes b$ Then, note that $\sum_{b\in B} b^\ast(w)b = w$ for all $w\in W$ - simply by the definition of the dual basis, so $\begin{array}{rcl} \phi(\psi(g))(v) &=& \sum_{b\in B} b^\ast(g(v))\cdot b = g(v) \\ \psi(\phi(f\otimes w))&=&\sum_{b\in B} (b^\ast\circ(v\mapsto f(v)w)\otimes b = \sum_{b\in B} (b^\ast(w)\cdot f)\otimes b \\ &=& f \otimes \sum_{b\in B} b^\ast(w)b = f\otimes w. \end{array}$ We can now extend this result to show that $V_1^\ast \otimes \cdots \otimes V_n^\ast \otimes W \cong L_n(V_1,\ldots,V_n; W),$ the space of multilinear maps $V_1\times\cdots\times V_n \to W$. In fact, this is best done by induction: Simply consider the (obvious) isomorphism $\begin{array}{rcl} L_{n-1}(V_1,\ldots,V_{n-1}; \mathrm{Hom}_k(V_n, W)) &\longrightarrow& L_n(V_1,\ldots,V_n; W)\\ f &\longrightarrow& ( (v_1,\ldots,v_n)\mapsto f(v_1,\ldots,v_{n-1})(v_n)) \\ g(v_1,\ldots,v_{n-1},\,\cdot\,) &\longleftarrow& g \end{array}$ and use it for the induction step.

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