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I'm trying to better understand the symmetric algebra and how it behaves on the dual space of the algebra. Wikipedia states the following:

In mathematics, the symmetric algebra $S(V)$... corresponds to polynomials with indeterminates in $V$, without choosing coordinates. The dual, $S(V^∗)$ corresponds to polynomials on $V$.

I'm trying to understand this distinction better. I think that what they're saying is that $S(V)$ is the set of formal polynomials taking indeterminates in $V$, whereas $S(V^*)$ can be identified with the set of polynomial functions on $V$. I would appreciate a sanity check that I am seeing this right.

Here's how I'm viewing it so far:

For example, suppose your vector space $V$ is over $\Bbb F_2$, the finite field of order 2. Then for any vector $v$, the polynomial $v^2 - v$ is well-defined as a formal expression, and hence is an element of the symmetric algebra $S(V)$.

On the other hand, let $f$ be a linear functional in the dual space $V*$. Then we can likewise construct a formal polynomial $f^2-f$. However, if we actually treat this as a polynomial function on $V$, then for any $v$, $f(v)^2-f(v) = 0$, and hence can be identified with the zero polynomial function.

In other words, in this interpretation, $v^2-v$ exists as a member of $S(V)$, as it is a formal polynomial. However, within $S(V*)$, the polynomial function $f^2-f=0$.

Is this the correct way to view the distinction? Or is it more subtle than this?

Mike Battaglia
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1 Answers1

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No, this is not correct; $S(V)$ and $S(V^{\ast})$ are both polynomial algebras, and each is exactly as formal as the other. In particular, $f^2 \neq f$.

Over a finite field "polynomials on $V$" needs to be interpreted more carefully: a (finite-dimensional) vector space over $k$ can naturally be upgraded into an affine (group) scheme over $k$, whose set of $k$-points is $V$, and the ring of functions on this affine scheme (not on $V$ itself) is $S(V^{\ast})$. The cleanest way to describe this scheme is to describe its functor of points: if $k$ is the base field, then the functor of points of $V$-as-an-affine-scheme sends a commutative $k$-algebra $A$ to

$$V \otimes_k A.$$

Then it's an exercise to show that this functor is represented by $S(V^{\ast})$.

Qiaochu Yuan
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  • OK, would you say this answer is incorrect then? https://math.stackexchange.com/questions/864896/natural-isomorphism-of-polynomial-functions-on-v-and-sv

    "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, ${\rm Sym}(V^*)$."

    "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." (cont'd)

     – Mike Battaglia May 30 '17 at 19:51
  • Basically, I see stuff like this pop up sometimes, identifying the ring of polynomial functions with $S(V^*)$, and I can't tell where the discrepancy comes from. Is stuff like this just wrong, or is it a different convention, or...? – Mike Battaglia May 30 '17 at 19:51
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    @Mike: it's correct if either the underlying field is infinite or if you adopt the convention above that "polynomials on $V$" refers to this more refined algebro-geometric thing. But it seems like you might also be confused about a different thing, which is the distinction between $S(V)$ and $S(V^{\ast})$. A basic distinction is that $S(V)$ is covariant and $S(V^{\ast})$ is contravariant; functions are contravariant because they pull back. – Qiaochu Yuan May 30 '17 at 19:52
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    OK, that makes sense. Thank you! – Mike Battaglia May 30 '17 at 22:54