## May 31, 2007

### What is a Lie Derivative, really?

#### Posted by Urs Schreiber

For solving the problem that I am currently working on, it turned out I need to understand

What is a Lie derivative, really?

By which I mean

What is a Lie derivative, arrow-theoretically?

By which I mean

How can I think of a Lie derivative in an implementation-independent way, such that the concept may be a) internalized and, in particular, b) be categorified without effort (read: without running into problems that require thinking).

As David Corfield has put it in The Two Cultures of Mathematics Revisited:

“[…] for any worthwhile idea there is a story about it which gets to the heart of what it really is, and I’ll know when I’ve reached that point by the ease with which it categorifies.”

And this may be necessary for understanding what’s going on.

So here is my current take at the answer to “What is a Lie derivative, really?”. It’s maybe not quite the final answer yet, but the applications that I am looking at suggest that this is on the right track.

Let me know what you think!

Question: What is a Lie derivative really?

Answer (somewhat preliminary): For any given object $C$ in a ($n \gt 1$)-category, Lie derivatives on $C$ are 2-morphisms starting at the identity morphism $\array{ & \nearrow \searrow^{\mathrm{Id}} \\ C &\Downarrow^f& C \\ & \searrow \nearrow_{t(f)} } \,.$ These compose using horizontal composition of transformations.

Examples:

1) ordinary Lie derivatives: Let $C = (\Omega^\bullet(X),d)$ be the differential graded commutative algebra of differential forms on a manifold $X$. This is an object in an $\infty$-category (I think) whose morphisms are degree 0-derivations, whose 2-morphisms are chain homotopies of these (hence in particular degree -1 derivations), etc.

Then a Lie derivative as a above is a chain homotopy $i_v$ (contraction with vector field $v \in \Gamma(T X)$) $\array{ & \nearrow \searrow^{0} \\ (\Omega^\bullet(X),d) &\Downarrow^{\iota_v}& (\Omega^\bullet(X),d) \\ & \searrow \nearrow_{L_v = [d,\iota_v]} } \,.$

2) the integrated version of that

The above example may be regarded as the Lie algebroid version of the following groupoid situation:

Let $P_1(X)$ be the path groupoid of $X$ (morphisms are thin-homotopy classes of paths in $X$).

In Isham on Arrow Fields I reviewed the discussion of From Arrows to Disks later extended to The Algebra of Observables, that flows along vector fields on $X$ should arrow-theoretically be thought of as transformations: $t \;\;\; \mapsto \;\;\; \array{ & \nearrow \searrow^{\mathrm{Id}} \\ P_1(X) &\Downarrow^{v}& P_1(X) \\ & \searrow \nearrow_{\exp(v)(t)} } \,.$ Here, more precisely, the flow is a smooth Lie group homomorphism from the additive group of real numbers to the group of of transformations under horizontal composition as above.

2) the puzzle posed in Derivation Lie 1-Algebras of Lie $n$-Algebras:

Every Lie $n$-algebra $g_{(n)}$ can equivalently be conceived as a free graded commutative algebra with a differential on it $(\wedge^\bullet (sV^*), d_{g_{(n)}})$ (described here).

This means we have the Lie derivatives $\array{ & \nearrow \searrow^{0} \\ (\wedge^\bullet sV^*, d_{g_{(n)}}) &\Downarrow^{\iota_x}& (\wedge^\bullet sV^*, d_{g_{(n)}}) \\ & \searrow \nearrow_{L_x = [d,\iota_x]} }$ for each degree -1 derivation $\iota_x$ on $g_{(n)}$.

I was asking for the integrated version of this. Following this argument I am pretty sure (but still haven’t tried to fully prove this in all detail), that for $G_{(n)}$ a Lie $n$-group integrating $g_{(n)}$, the corresponding Lie group is that of the transformations

$\array{ & \nearrow \searrow^{\mathrm{Id}} \\ \Sigma(G_{(n)}) &\Downarrow^{f}& \Sigma(G_{(n)}) \\ & \searrow \nearrow_{f(f)} } \,.$

There is a story to be told how these three examples taken together help to find the answer to

What is a Cartan connection, really?

What is a $g_{(n)}$-connection , for $g_{(n)}$ any Lie $n$-algebra?

But that story will be told elsehere.

Posted at May 31, 2007 10:50 AM UTC

TrackBack URL for this Entry:   http://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/1292

### Re: What is a Lie derivative, really?

This is really interesting Urs! I think I’m starting…slowly… to understand what you and Jim Stasheff have been talking about.

Posted by: Bruce Bartlett on May 31, 2007 1:13 PM | Permalink | Reply to this

### Re: What is a Lie derivative, really?

This is really interesting Urs!

Thanks!

Alas, in real life it is not quite sufficient for something to be interesting – what matters, too, is that people find it interesting.

Therefore I am glad that you do! :-)

I’m starting…slowly.. to understand what you and Jim Stasheff have been talking about.

It might be somewhat hard to discern, because the main point hasn’t really been revealed yet. I hinted at that at the end of the above entry. Am busy preparing a discussion of that currently concealed part.

On the other hand, I am miles away from the brilliant exposition that John Baez cultivates. So things may appear more opaque than they are.

Therefore I ask you: please ask questions if you find this interesting but hard to grasp. There is nothing in here which you wouldn’t easily understand, so if you don’t, it’s my fault. Please ask. Stepwise explanation will make it all clear, I think.

Posted by: urs on May 31, 2007 1:25 PM | Permalink | Reply to this

### Differentiation as Functor; Re: What is a Lie derivative, really?

Does this properly generalize the very short paper (and is that paper’s proof sketch correct):

“Mathematical Minutiae: Differentiation as a Functor”

by Athanasios Papioannou, Harvard ‘07

Unlike any other article in this journal, this one begins with a warning: Categories, beautiful and powerful as they may be, are not panacea and should be used with great prudence. This short note presents a fun, but silly use of categories.

Harvard College Mathematics Review, Vol. 1, No. 1, Spring 2007, p.77.

http://www.hcs.harvard.edu/hcmr/issue1/thanos.pdf

Posted by: Jonathan Vos Post on May 31, 2007 8:45 PM | Permalink | Reply to this

### Re: Differentiation as Functor; Re: What is a Lie derivative, really?

Harvard College Mathematics Review, Vol. 1, No. 1, Spring 2007, p.77.

http://www.hcs.harvard.edu/hcmr/issue1/thanos.pdf

Strange document that. It’s true that differentiation $T : C^\infty \to C^\infty$ $T : (X \stackrel{f}{\to} Y) \mapsto (T X \stackrel{d f}{\to} T Y)$ is functorial.

Just see the Wikipedia article on the Chain rule.

Not sure in which sense an elementary fact of this sort can be “silly”.

Posted by: urs on May 31, 2007 8:59 PM | Permalink | Reply to this
Read the post The second Edge of the Cube
Weblog: The n-Category Café
Excerpt: Differentiating parallel transport anafunctors to Cartan-Ehresmann connections.
Tracked: May 31, 2007 9:44 PM

### What is a Lie Derivative, really?

A map between modules of tensor fields which commutes with the action of the diffeomorphism group.

In situations where only a subgroup are relevant, e.g. on symplectic or contact manifolds, substitute this subgroup for the diffeomorphism group above.

### Re: What is a Lie Derivative, really?

My brain evidently took summer vacation early this year - for some reason I read covariant derivative instead of Lie derivative.

The Lie derivative is conceptually even simpler: it is a representation of the diffeomorphism algebra. This is clear, because the Lie derivative LX satisfies

[LX, LY] = L[X,Y].

This is not only a conceptually clear viewpoint but also a fruitful one, because it leads to a vast generalization of the concept of Lie derivative: instead of just considering proper reps of the diffeomorphism algebra, consider projective ones. On the circle, this generalization takes us from primary and secondary fields to lowest-weight reps of the Virasoro algebra.

### Re: What is a Lie Derivative, really?

But not just for the diffeo algebra
nor even just sub algs thereof
It has a purely algebraic defintion in
the H Cartan formalism

jim

Posted by: jim stasheff on June 8, 2007 3:01 PM | Permalink | Reply to this

### Re: What is a Lie Derivative, really?

I’ll say it again:

given $(A,d)$ any graded differential algebra with differential $d$ of degree +1 and $d^2 = 0$, and $\tau : A \to A$ any degree -1 derivation, we get a degree 0 derivation $L_\tau = [d,\tau] \,.$ This is the Lie derivative associated with $\tau$.

These Lie derivatives (and hence the degree -1 derivations) always form a Lie algebra: for any such degree -1 derivation $\tau_1$ and $\tau_2$ there is a degree -1 derivation $\tau_3$ such that $[L_{\tau_1},L_{\tau_2}] = L_{\tau_3} \,.$ This follows directly from $d^2 = 0$ and the graded Jacobi identity.

Posted by: urs on June 8, 2007 3:05 PM | Permalink | Reply to this

### Re: What is a Lie Derivative, really?

Hi Urs,

Could you explain a bit more about what cartan connections are, and what they’re good for? I tried to read the wikipedia article but it’s quite a long one and I’m not sure which is the way you would encourage us to think of them.

Posted by: Bruce Bartlett on June 7, 2007 12:43 AM | Permalink | Reply to this

### Re: What is a Lie Derivative, really?

Bruce,

You win the prize for making the $5000^{th}$ comment on this blog. Help yourself to free virtual coffee and brownies all day.

Posted by: David Corfield on June 7, 2007 9:38 AM | Permalink | Reply to this

### Re: What is a Lie Derivative, really?

Hooray! Congratulations guys!

Posted by: Bruce Bartlett on June 7, 2007 9:52 AM | Permalink | Reply to this

### Re: What is a Lie Derivative, really?

Could you explain a bit more about what cartan connections are

Sure! I have now posted a reply here.

Posted by: urs on June 11, 2007 2:09 PM | Permalink | Reply to this
Read the post Polyvector Super-Poincaré Algebras
Weblog: The n-Category Café
Excerpt: Superextension of Poincare algebras and how these give rise to brane charges.
Tracked: June 14, 2007 5:15 PM