2
$\begingroup$

I'm reading a paper and I didn't understand this notation used by the author:

Let E be a vector space and F be a subspace of E. Let $S(E/F)$ be the symmetric algebra of $E/F$. For every element $P \in S(E/F)$, we denote $P(\partial/ \partial {X_2}|_0)$ the application:

$ P(\frac{\partial}{\partial {X_2}}|_0) : C^\infty(E)\longrightarrow C^\infty(F) $

Define for $f \in C^\infty(E)$ by

$(P(\frac{\partial}{\partial {X_2}}|_0)f)(X_1)= (P(\frac{\partial}{\partial {X_2}})f)(X_1 + X_2)|_{x_2 = 0} $

$X_1 \in F, X_2 \in E/F$.

My question is what is the application $P(\frac{\partial}{\partial {X_2}})f $ (how it is defined ?) which is used by the author to define the application $ P(\frac{\partial}{\partial {X_2}}|_0) $ ?

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ The notation $f(X_1+X_2)$ suggests that there is an implicit isomorphism $E \cong F + E/F$ in play. Via the isomorphism, $X_1$ and $X_2$ are elements of $E$ and so $P(\frac{\partial}{\partial X_2})$ has its literal meaning. The question is whether the result depends on the isomorphism, it being non-canonical. I think No. Another way to understand $P(\frac{\partial}{\partial X}) f|_{X=0}$ is as the natural pairing of $P\in S(E)$ with the Taylor polynomial of $f$ in $S(E^*)$. But this pairing descends to the one between $S(E/F)$ and $S(F^\perp)$ in a way independent of the isomorphism. $\endgroup$
    – Igor Khavkine
    Dec 7, 2020 at 22:55
  • $\begingroup$ Thank you so much for your comment ! Could you please explain to me what is the pairing of $ P \in S(E)$ and the taylor polynomial of f in $S(E^*)$ by giving a simple example or pointing me to a reference where I can find it. $\endgroup$
    – Maria
    Dec 7, 2020 at 23:39
  • $\begingroup$ @Igor Khavkine if we denote $\langle . , .\rangle $ this pairing and denote T(f) the taylor exapansion of f at specific order , then is this true $p (\frac{\partial}{\partial X })f := \langle P, \frac{\partial}{\partial X } T(f) \rangle$ ? $\endgroup$
    – Maria
    Dec 7, 2020 at 23:54
  • 1
    $\begingroup$ yes up to some constant, and actually just $p(\frac{\partial}{\partial X}) f|_{X=0} = \langle P, T(f) \rangle$. This is elementary, and you can easily check it in one dimension: $\partial^k_x x^k |_{x=0} = k! = k! \langle \partial^k_x, x^k \rangle$. $\endgroup$
    – Igor Khavkine
    Dec 8, 2020 at 0:28
  • $\begingroup$ @Igor Khavkine thank you so much! $\endgroup$
    – Maria
    Dec 8, 2020 at 0:47

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.