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In Humphreys, reflection groups and coxeter group book, Humphreys denotes $S(V^*)$ as the ring of polynomial function on the finite dim vector space $V$. Why we are considering $S(V^*)$ rather than $S(V)$?.

Can somebody explains me this clearly?

Thanks in Advance.

GA316
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If you want polynomial expressions whose variables are vectors of $V$, then the full symmetric algebra ${\rm Sym}(V)$ is the way to go: this allows you to form and then multiply arbitrary monomials of vectors, linearly combine them, and distribute addition of vectors and more complicated expressions over multiplication, all exactly as desired.

If you want polynomial functions $V\to F$, that's a different story. Suppose $f:V\to F$ is a function from the space to its underlying field, and suppose we choose a basis. Every vector is then a unique linear combination of basis vectors - essentially, a coordinate vector. The function is polynomial if it is expressible as a polynomial in these coordinates. (For instance, $(a,b,c)\mapsto ab+c^2$.) One can easily prove that a function is polynomial with respect to one basis if and only if it is polynomial with respect to all bases, so being polynomial is not basis-dependent.

The maps $(a,b,c,\cdots)\mapsto a$, $(a,b,c,\cdots)\mapsto b$, $(a,b,c,\cdots)\mapsto c$, $\cdots$ are all instances of linear functionals $V\to F$. Every polynomial function is a polynomial in these linear functionals, and the space of linear functionals is precisely $V^*$. Hence the algebra of polynomial functions can be identified with the full symmetric algebra of the dual space, ${\rm Sym}(V^*)$.

A monomial $w_1\cdots w_k\in{\rm Sym}(V^*)$ acts as a function on $V$ by $v\mapsto w_1(v)\cdots w_k(v)$.

anon
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  • very nice. but I am still not clear. what is the advantage or naturality of $S(V^*)$ over $S(V)$.Whatever you said above can be done in $S(V)$ also no? please explain little bit more clearly. – GA316 Jul 12 '14 at 09:15
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    @GA316 How would you interpret an element of $S(V)$ as a function on $V$? In particular, given $V\subseteq S(V)$, how would you interpret vectors in $V$ as linear functionals on $V$? That's tantamount to specifying an isomorphism $V\cong V^$, and there is no *natural isomorphism between a vector space and its dual. The advantage here is that $S(V^*)$ is naturally identified as the algebra of polynomial functions, but there is no natural way to do that for $S(V)$ at all. – anon Jul 12 '14 at 09:25
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    To put it in as clear of terms as possible: $S(V^*)$ is the algebra of polynomial functions on $V$, and $S(V)$ isn't, and that's all there is to it. – anon Jul 12 '14 at 09:34