Veronese embedding of degree $d$ maps $\mathbb P^n$ to $\mathbb P^m$, where $m =\binom{n+d}{n} -1$, and is commonly given by the formula $\nu: \left(x_0:x_1:\dots:x_n\right) \mapsto \left(x_0^d: x_0^{d-1} x_1: \dots : x_n^d\right)$.
I see two possible coordinate-free interpretation of the formula with seemingly different target spaces. The first is simply $V \to Sym^d V: v \mapsto v^d$. The second is $V \to Sym^d \left ( V^*\right)^*$, since the evaluation of polynomial $\sum a_{i_0,\dots,i_n} x_0^{i_0}\dots x_n^{i_n}$ on $v \in V$ is linear in polynomial coefficients. It thus can be seen as pairing between $\left(a_{i_0,\dots,i_n}\right) \in Sym^d V^*$ and $\nu(v).$
That raises a question: is there a natural way to relate $Sym^d V$ and $Sym^d \left(V^* \right)^*$?

  • $\begingroup$ There is a canonical isomorphism $V \cong (V^*)^*$ that induces canonical isomorphisms on all tensor constructions. $\endgroup$ – Tabes Bridges Apr 10 '17 at 18:04
  • $\begingroup$ @tabes Well, for some reason I was not sure that taking dual space commutes with symmetric power. I think I see this now. $\endgroup$ – Troshkin Michael Apr 10 '17 at 19:20
  • $\begingroup$ Oh, I see; I was reading your notation incorrectly. You're saying dualize $V$, take the symmetric power, and dualize again. Off hand I'm unsure if this is canonical. $\endgroup$ – Tabes Bridges Apr 11 '17 at 17:24
  • $\begingroup$ @tabes Well, $T^k \left( V \right)$ and $T^k \left( V^* \right)$ are dual with rather obvious pairing given by $\left \langle v_1 \otimes \dots \otimes v_k, \xi_1 \otimes \dots \otimes \xi_k\right \rangle = \prod_i \left \langle v_i, \xi_i\right \rangle$ for $v_i \in V$, $\xi_i \in V^*$. However, this pairing is not preserved under permutations of tensor components unless all components are identical. This is what happens in my question: a map $W \to Sym^d\left( V^* \right)^*$ can be defined for a subspace $W$ in $T^d V$ or $Sym^d V$ generated by tensors $v^d$ for $v \in V$. $\endgroup$ – Troshkin Michael Apr 12 '17 at 20:27
  • $\begingroup$ However, $W \subset Sym^d V$ must be $Sym^d V$ itself as Veronese variety is not contained in any hyperplane (see math.stackexchange.com/questions/19950/…, for example). Consequently, $W' \subset T^d V$ is the subspace of symmetric tensors, which allows us to see that every symmetric tensor decomposes into sum of d-th powers. $\endgroup$ – Troshkin Michael Apr 12 '17 at 21:13

I have managed to figure out a very good candidate for natural pairing between $Sym^d V$ and $Sym^d ( V^* )$. There is a duality between $T^d V$ and $T^d( V^*)$, defined by $$\left \langle v_1 \otimes \dots \otimes v_d, \xi_1 \otimes \dots \otimes \xi_d \right \rangle = \prod_i \left \langle v_i, \xi_i \right \rangle $$for $v_i \in V,\; \xi_i \in V^* $.
Let $ \sigma: Sym^d V \to ST^d V \subset T^dV $ and $\sigma^*:Sym^d(V^*) \to ST^d(V^*) $ be natural isomorhisms between symmetric powers and spaces of symmetric tensors. Then duality between $Sym^d V$ and $Sym^d(V^*)$ is defined as $\left \langle v,\; \xi \right \rangle = \left \langle \sigma(v), \; \sigma^*( \xi) \right \rangle$ for $v \in Sym^d V$, $\xi \in Sym^d(V^*)$.
As one might observe, this definition agrees with two Veronese mappings described in the question, creating a commutative triangle of $V,\; Sym^d V$ and $Sym^d(V^*)^*$ as well as corresponding projective spaces.

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