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Let $K$ be a field and $V$ be a linear space over $K$. A map $p\colon V \to K$ is homogeneous polynomial of degree $n$ if there exist the symmetric $n$-linear form $f\colon V^{\times n}\to K$ such that $p(x)=f(x,x,\ldots,x)$ for any $x\in V$.

The question: where (what books or articles?) can I read about the map $\delta_a p\colon V \to K$, $a\in V$, such that $\delta_a p(x)=nf(a,x,\ldots,x)$ for any $x\in V$? As far as I understand, the map is the derivative of $p$ along $a$, is it? So I probably need some papers related to differential geometry?

1 Answers 1

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What do you actually need to know about the map?

Consider the partial derivative in the $a$ direction of $p$:

$ \nabla_a p(x) = \lim_{t\to 0} \frac{p(x+ta) - p(x)}{t} $

and note that

$ p(x+ta) = f(x+ta, x+ta, \ldots, x+ta) $

which you can expand by linearity. (I'll show you the case where $n = 3$, you can try to do the general case yourself.)

$ \begin{align} f(x+ta,x+ta,x+ta) &= f(x+ta,x+ta,x) + tf(x+ta,x+ta,a) \\ &= f(x+ta,x,x) + tf(x+ta,a,x) + tf(x+ta,x,a) + t^2f(x+ta,a,a) \\ &= f(x,x,x) + t \left(f(x,x,a) + f(x,a,x) + f(a,x,x)\right) \\ &\quad + t^2 \left(f(x,a,a) + f(a,x,a) + f(a,a,x)\right) + t^3 f(a,a,a)\end{align}$

which you can simplify, using the fact that $f$ is symmetric, to

$ f(x,x,x) + 3t f(x,x,a) + 3t^2 f(x,a,a) + f(a,a,a) $

In general, using linearity and symmetry you can show that a binomial-theorem like formula must hold for $p(x+ta)$. And using this, you indeed get

$\nabla_a p(x) = n f(a,x,x,\ldots,x)$