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The following is mildly edited from the beginning of Bhargava's Higher Composition Laws I:

We will denote by $(\text{Sym}^k \mathbb{Z}^n)^*$ the lattice of $n$-ary $k$-ic forms with integer coefficients. The reason for the * is that there is also a sublattice $\text{Sym}^k \mathbb{Z}^n$ corresponding to the forms $f: \mathbb{Z}^n \rightarrow \mathbb{Z}$ satisfying $f(\xi) = F(\xi, \dots, \xi)$ for some symmetric multilinear function $F: \mathbb{Z}^n \times \cdots \mathbb{Z}^n \rightarrow \mathbb{Z}$ (classically called the `polarization' of $f$). Thus, for example, $(\text{Sym}^2 \mathbb{Z}^2)^*$ is the space of binary quadratic forms $f(x, y) = ax^2 + bxy + cy^2$ with $a, b, c \in \mathbb{Z}$, while $\text{Sym}^2 \mathbb{Z}^2$ is the subspace of such forms where $b$ is even.

This is enough to read and appreciate Bhargava's paper. That said, it seems that this should follow from standard definitions in representation theory, but I couldn't find these thoroughly treated in any of the reference books I have (e.g. Fulton-Harris).

It is possible to make some observations, for example, that $(\text{Sym}^k \mathbb{Z}^n)^*$ can naturally be constructed as a quotient of the tensor algebra, where $\text{Sym}^k \mathbb{Z}^n$ is a subspace. But it also seems to be possible to ask questions whose answers aren't obvious. For example, is the double dual of $\text{Sym}^k \mathbb{Z}^n$ isomorphic to $\text{Sym}^k \mathbb{Z}^n$? As abelian groups, sure, but working in the category of abelian groups would not preserve the distinction between $\text{Sym}^k \mathbb{Z}^n$ and its dual.

What are the intrinsic definitions, from which (I would guess) I could easily verify the above facts and answer my own question? Are these treated in some reference which I have overlooked?

Thank you.

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  • $\begingroup$ I've always seen $\operatorname{Sym}^k$ defined as a quotient, not as a subspace. Moreover isn't it more natural to consider $\operatorname{Sym}^k(({\bf Z}^n)^*)$? To put this in bijection with the $n$-ary $k$-ic forms, we map $\lambda_1 \otimes \cdots \otimes \lambda_k$ to the form $x \mapsto \lambda_1(x) \cdots \lambda_k(x)$. $\endgroup$ Jan 27, 2013 at 15:30
  • $\begingroup$ Duals of symmetric powers of a finitely generated projective module are the graded parts of the divided power algebra of the dual module, a different creature from the symmetric powers of the dual module when working over a base like $\mathbf{Z}$ that isn't a $\mathbf{Q}$-algebra. So blindly mixing up submodules and quotients through duality can create problems over $\mathbf{Z}$ in a way that one doesn't notice when working over $\mathbf{Q}$. This highlights the significance of Francois' comment (where the dual occurs inside the symmetric power). $\endgroup$
    – user30379
    Jan 27, 2013 at 15:43
  • 1
    $\begingroup$ See mathoverflow.net/questions/34452 $\endgroup$ Jan 27, 2013 at 16:33
  • 1
    $\begingroup$ See mathoverflow.net/questions/15336 $\endgroup$ Jan 27, 2013 at 16:34
  • 1
    $\begingroup$ See mathoverflow.net/questions/45664 $\endgroup$ Jan 27, 2013 at 16:35

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