# A dual space to a space of homogeneous polynomials

I stumbled across the following issue. Let say we have an $$n$$-dimensional vector space $$V\simeq \mathbb{F}^{n}$$ and let $$V^{*}$$ be its dual. There is no natural way to construct a natural isomorphism between them without choosing a specific basis. Let $$\{e^{1},e^{2},\dots,e^{n}\}$$ be some basis of $$V$$. Then we can define the dual $$\{e_{1},e_{2},\dots,e_{n}\}$$ basis for $$V^{*}$$ such that $$e_{i}(e^{j}) = \delta_{i,j}$$. There is a way to make it more tangible. Set an isomorphism $$V \simeq\mathbb{F}^{n}$$ by identifying $$e^{i}$$ with $$\begin{pmatrix}0&0&\cdots&1&\cdots&0\end{pmatrix}^{\mathsf{T}}$$ a column with $$1$$ at the $$i$$-th position. Then $$e_{j}$$ can be identified with $$\begin{pmatrix}0&0&\cdots&1&\cdots&0\end{pmatrix}$$ a row with $$1$$ at the $$j$$-th position and $$e_{i}(e^{j})$$ is the matrix multiplication of the row with a column. This is particularly nice because it allows us to write all elements of $$V^{*}$$ as rows with natural addition and multiplication by numbers.

Now let's realize $$V$$ in a different manner, particularly as $$S^{n - 1}(\mathbb{F}x\oplus\mathbb{F}y)$$. In other words $$V$$ is isomorphic to a space of all homogenous polynomials of degree $$n-1$$ in variables $$x$$ and $$y$$: $$V\simeq \left\{\sum_{i = 0}^{n-1}a_{i}x^{n-1-i}y^{i}\mathrel{\Big|} a_{i}\in\mathbb{F}\right\}$$ We can construct the dual of this with a use of substitutions. For example we can define $$\varphi_{(x_{0},y_{0})}\in V^{*}$$ mapping a polynomial $$P(x,y)\in V$$ to $$P(x_{0},y_{0})$$. But I don't see how the addition and multiplication by the elements of $$\mathbb{F}$$ can be defined except formally and I don't understand how we can define a more or less natural dual basis in this case.

Are there any other natural ways to do it? Can we find a basis which nicely behaves under multiplication in graded algebra $$S(\mathbb{F}x\oplus\mathbb{F}y)$$? I'm interested because it's an important construction in representation theory of matrix groups.

The dual can be described in terms of differential operators. Given a polynomial $$p(x, y)$$ in two variables we can consider the family $$f_{i, j}$$ of linear functionals defined by taking the partial derivative $$\frac{ \partial_x^i}{i!} \frac{\partial_y^j}{j!}$$ (note that this makes sense in arbitrary characteristic), then evaluating at $$x = y = 0$$. This just computes the Taylor series expansion of $$p$$ at the origin, so if $$p(x, y) = \sum p_{i, j} x^i y^j$$ we just have

$$f_{i, j}(p) = p_{i, j}.$$

So for fixed $$n$$ the linear functionals $$f_{i, j}, i + j = n$$ are a basis of the dual of the homogeneous polynomials of degree $$n$$, namely the dual basis of the basis of monomials.

Of course we could have said "dual basis of the basis of monomials" without phrasing this in terms of derivatives. The reason to phrase things in terms of derivatives is that there is more structure available. Namely, for $$V$$ a finite-dimensional vector space, the entire symmetric algebra $$S(V)$$ is naturally a graded Hopf algebra with comultiplication given by extending

$$\Delta(v) = v \otimes 1 + 1 \otimes v.$$

This can be thought of either as the universal enveloping algebra $$U(V)$$ of $$V$$ regarded as an abelian Lie algebra (the "cocommutative picture"), or as the Hopf algebra of functions on the dual $$V^{\ast}$$ regarded as an affine group scheme (the "commutative picture"). It is both commutative and cocommutative. Now, a graded Hopf algebra whose graded pieces are finite-dimensional has a graded dual which is also a Hopf algebra with the same property, and taking the graded dual of $$S(V)$$ produces another Hopf algebra, which is also commutative and cocommutative, and with the same graded dimension.

Now a natural question arises: what is this Hopf algebra? The answer is that in characteristic $$0$$ it can be identified with $$S(V^{\ast})$$, and explicitly if we choose a basis $$x_1, \dots x_n$$ of $$V$$, then it can be identified with the algebra of differential operators $$K[\partial_1, \dots \partial_n]$$ generated by partial derivatives with respect to each of these. However, in positive characteristic it is genuinely different! A uniform description which works in all characteristics is that it is a free divided power algebra $$\Gamma(V^{\ast})$$. After choosing a basis $$x_1, \dots x_n$$ of $$V$$, so that we can think of $$S(V)$$ as the polynomial algebra $$K[x_1, \dots x_n]$$, we can think of the graded dual $$\Gamma(V^{\ast})$$ as the divided power algebra generated by the partial derivatives $$\partial_1, \dots \partial_n$$; here "divided powers" is a fancy way to say that we have the freedom not only to multiply the partial derivatives but to compute $$\frac{\partial_i^n}{n!}$$ and so forth, and this makes sense in arbitrary characteristic.

Specializing to a fixed degree, the upshot is that in characteristic zero the dual $$S^n(V)^{\ast}$$ can be identified with $$S^n(V^{\ast})$$. However, as discussed e.g. here, this identification turns out to involve dividing by $$n!$$, so it fails in characteristic $$p \le n$$. In these characteristics the two are genuinely different (I think even as representations of $$GL(V)$$) and we just get a different thing $$\Gamma^n(V^{\ast})$$ which can be described in terms of divided powers as above. I believe but have not checked carefully that an alternate description is that $$\Gamma^n(V^{\ast})$$ is the subspace $$\text{Sym}^n(V^{\ast})$$ of symmetric tensors in $$T^n(V^{\ast})$$.

Edit: Here's a quick way to see the identification with symmetric tensors. We just dualize the quotient map $$V^{\otimes n} \to S^n(V)$$ given by taking coinvariants with respect to the $$S_n$$ action, and we get the inclusion

$$\Gamma^n(V^{\ast}) \cong \text{Sym}^n(V^{\ast}) \to (V^{\ast})^{\otimes n}$$

given by taking invariants with respect to the $$S_n$$ action. In characteristic $$0$$ or $$p > n$$ symmetric tensors and the symmetric power (the invariants and coinvariants of the $$S_n$$ action) are canonically isomorphic but the canonical isomorphism involves dividing by $$n!$$ (and this is true more generally for any finite group $$G$$ acting on a finite-dimensional representation $$V$$, that invariants $$V^G$$ and coinvariants $$V_G$$ are canonically isomorphic if we can divide by $$|G|$$).

Here's also some more details on the relationship between differential operators and $$V^{\ast}$$. On the symmetric algebra $$S(V)$$, any element of the dual $$\partial \in V^{\ast}$$ extends uniquely to a derivation $$\partial : S(V) \to S(V)$$ of degree $$-1$$; if we think of $$S(V)$$ as polynomial functions on $$V^{\ast}$$ these are partial derivatives with respect to directions in $$V^{\ast}$$. These derivations commute, which is a quick way to see that we get a (natural!) action of $$S(V^{\ast})$$ on $$S(V)$$ by differential operators. But we get slightly more because of divided powers.

• Wow!! Thanks for such an extensive answer! There is a lot to digest here, but it all indeed is relevant for my studies both in $\mathbb{C}$ and $\mathbb{F}_{q}$ case! Commented Jul 22 at 21:02