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I have a trivial question concerning Lawson/Michelsohn's "Spin Geometry", Chapter III.§1. There, the symbol of a differential operator $P$ is defined to be a section $\sigma(P)$ in the bundle $(\odot^m TX) \otimes Hom(E,F)$. In the paragraph below, the following is claimed: "Recall that for a vector space $V$, the space $\odot^m V$ is canonically isomorphic to the space of homogenous polynomial functions of degree $m$ on $V^*$."

Question 1: How is the space $P^m(V^*)$ of "homogenous polynomial functions of degree $m$ on $V^*$ defined"?

A comment to a smiliar question suggests to me that one should define $P^m(V^*) := \odot^m V$. But then this claim in Lawson/Michelsohn would be an empty statement. That does not look reasonable to me.

Question 2: Am I assuming correctly that once this issue is resolved, the isomorphism should pop out of the universal property of the symmetric product as the map $f:\odot^m V \to P^m(V^*)$ induced by the map $F:V^m \to P^m(V^*)$ defined by setting

$$ F(v_1, \ldots, v_m)(\varphi_1, \ldots, \varphi_m) := \sum_{|\alpha|=m}{\varphi_1(v_1)^{\alpha_1} \ldots \varphi_m(v_m)^{\alpha_m}}, $$

where $(v_1, \ldots, v_m) \in V^m$, $(\varphi_1, \ldots, \varphi_m) \in (V^*)^m$?

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1 Answer 1

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You can define $P^m(V^*)$ directly. Let $\alpha(x)$ be a polynomial of degree $m$ in the variables $x_1, \dots x_n$. If we decompose a vector $v \in V^*$ in some basis, we can plug the coefficients of $v$ into $\alpha$ to obtain a number. Thus we have a function on $V^*$. This is a standard procedure of turning polynomials (as abstract objects) into functions by evaluation. Alternatively you could consider $P^m(V^*)$ to be a subspace of all functions on $V^*$ that have a polynomial representation in some (and so all) basis.

I think your second observation is correct.

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Thanks, I think this resolves my confusion. –  Meneldur Feb 13 '13 at 12:52

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