A quickie :) Harris writes that the vector space of d-homogeneous polynomials over $V$ are isomorphic to $\mathrm{Sym}^d(V^*)$, but I couldn't find the definition so far. How is it defined?
What is $Sym^d(V)$?
1
$\begingroup$
linear-algebra
algebraic-geometry
-
2$\operatorname{Sym}^{d}(V^{\ast})$ is [the $d$-th symmetric power of $V^{\ast}$](http://en.wikipedia.org/wiki/Symmetric_algebra) – 2011-05-02
1 Answers
4
Since Wikipedia seems to say it all, I'm making my comment into an answer. See symmetric algebra and in particular the section on interpretation as polynomials. The notation $\operatorname{Sym}^d (V^{\ast})$ denotes the elements of degree $d$ of the symmetric algebra over $V^{\ast}$.
-
0Another thing: the article makes distinction between elements of symmetric algebra and symmetric tensors, is the difference only in multiplication of the respective algebras? – 2011-05-02
-
1@Alexei: No, the main point is that the symmetric algebra are a *quotient algebra* of of the tensor algebra, while the symmetric tensors only form a *subspace* of the tensor algebra. The situation is quite analogous as with the (possibly) more familiar [exterior algebra](http://en.wikipedia.org/wiki/Exterior_algebra). – 2011-05-02
-
1@Alexei Averchenko In characteristic $0$, there is a cannonical isomorphism of vector spaces between the symmetric tensors and the symmetric algebra. The symmetric tensors are not closed under multiplication, and so there isn't really multiplication for them (so you can't say the multiplications are different). However, in positive characteristic, you no longer have an isomorphism. You can see this because composing the natural map from $\operatorname{Sym}_d(V)$ to symmetric tensors (taking the sum of all rearangements) with the natural quotient map multiplies by $d!$. – 2011-05-02