# Polarisation on tensor products

Theorem: Let $V, W$ be finite dimensional $\mathbb{C}$-vectorspaces.

If $\alpha: V^{\otimes N} \to W$ is linear, then $\beta: V \to W, x \mapsto \alpha(x^{\otimes N})$ is homogeneous of degree $N$. Every homogeneous map $V \to W$ of degree $N$ is obtained this way.

While I do understand, why the first part is true ($\beta$ is homogeneous), the secound part is unclear to me (Why every hom. $V \to W$ is obtained this way). In our lecture, we didn't prove this, so maybe its trivial, but I don't see it.

• It should be $f(r \cdot v ) = r^n f(v)$, since this is precisely what $\beta$ does. – pepa.dvorak Oct 18 '17 at 10:37
Given a homogeneous $f: V \rightarrow W$, you're in fact looking for a homomorphism $\alpha: V^{\otimes n} \rightarrow W$ with $f = \alpha \beta$. Any such $\alpha$ is defined by its images on a basis of $V^{\otimes n}$, so let's define it just partially on elements of the form $v^{\otimes n}$ as $\alpha (v^{\otimes n}) = f (v)$ and let $\alpha$ be any extension of this partial homomorphism. Then you're done.
(A possible question could be: "does there exist some such extension?". There does - as $V^{\otimes n}$ is a vector space, any map defined by images an some basis (and extended linearly) is a homomorphism. Now, take a basis which contains the subset $\{v^{\otimes n} | \rm{v \, is \,an \,element \,of \,a \,basis\, of \,V}\}$.