4
$\begingroup$

There are more general definitions, but for my purposes a Lie algebroid on a smooth manifold $X$ is a vector bundle $A \to X$, a map $\rho: A \to {\rm T}X$ of vector bundles over $X$, and a bracket $[,]$ making $\Gamma(A)$ into an $\mathbb R$-Lie algebra, such that $\rho$ induces is a map $\Gamma(A) \to \Gamma({\rm T}X)$ of Lie algebras and such that the Leibniz rule $[a,fb] = f[a,b] + (\rho(a)f)b$ is satisfied for $f\in \mathcal C^\infty$ and $a,b\in \Gamma(A)$.

So, suppose I happen to have a Lie algebroid $A\to X$ lying around, and also a smooth map $\phi: Y\to X$. (In my case, $Y \to X$ happens to be a vector bundle, so you can assume some fairly strong properties of the map.) Then I can certainly pull back the vector bundle $A\to X$ to $\phi^*A \to Y$. Is there a natural Lie algebroid structure I can put on this pullback? The answer is probably "how natural do you want it?": if $Y = \{{\rm pt}\}$ and $\phi({\rm pt}) = y\in X$, then the anchor map for $\phi^*A \to \{{\rm pt}\}$ must be trivial, but the only part of the fiber $A_y$ with a Lie algebra structure is $\ker \{A_y \overset\rho\to {\rm T}_y X\}$.

So the real question is:

Along what type of maps do Lie algebroids pull back? In particular, can I always pull back a Lie algebroid along a submersion?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
6
$\begingroup$

As your example suggests, the vector bundle pullback is perhaps not the right thing to consider when wanting defining the notion of a pullback Lie algebroid. You have to go one step further and take the pullback of $\phi^*\rho: \phi^* A \to \phi^*TX$ along $d\phi: TY \to \phi^*TX$. The resulting bundle $\phi^{**}A$ (which exists when $d\phi - \phi^*\rho: TY \oplus \phi^*A \to \phi^*TX$ has constant rank, so in particular when $\phi$ is a submersion) then has a Lie algebroid structure with gives the right universal property you would want, if that's what you want. In the case where $Y$ is a point, this gives exactly your example.

This is all detailed in the paper "Algebraic Constructions in the Category of Lie Algebroids" by Higgins and Mackenzie. This may even have something to say about just trying to do something with $\phi^*A$ itself, but I haven't checked.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Awesome. In fact, awesomer than you know. At least, I'm trying to build a Lie algebroid that in the trivialized case comes in pieces, one of which is a Lie algebroid on $\phi^*A \to Y$. (Which is to say: I know what I want when $A\to X$ and $Y\to X$ are both trivialized bundles, but it's not obviously invariant, so I can't immediately glue.) But a dimension count suggests that this double pullback $\phi^{**}A$ has exactly the correct dimension to be the Lie algebroid I'm trying to construct. So now I just have to check if it really is. $\endgroup$
    – Theo Johnson-Freyd
    Commented May 31, 2010 at 1:06
  • $\begingroup$ In particular, since in Theo's case $Y \to X$ was a vector bundle, then $phi^{**} A \to Y$ is the ``total space'' of a VB-algebroid $(\phi^{**}A, Y, A, X)$. See section 6.1 on arXiv:0810.0066. Not sure whether this is relevant, but since you mention "in trivialized case comes in pieces", it may. $\endgroup$
    – Alfonso Gracia-Saz
    Commented Jun 8, 2010 at 2:09
  • $\begingroup$ It is also mentioned in the book "general theory of Lie groupoids and Lie algebroids" Page 156 (in the hardcopy thag i have).. $\endgroup$
    – Praphulla Koushik
    Commented Dec 11, 2021 at 12:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .