Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let M be an m dimensional differentiable manifold. Define Gauge(M):=C^{\infty}(M, Aut(TM)) to be the group of all (smooth) fiberwise linear transformations of the tangent bundle. This is the natural gauge group of the manifold. If (U, x_1,...,x_m) is a local coordinate system with induced frame on TU then an element of Gauge(U) looks like an invetable matrix g_{ij}(x_1,...,x_m) (with i,j=1,...,m) depending smoothly on the point. If we take a diffeomorphism of M interpreted as a coordinate transformation i.e., taking (U,x_1,...,x_m) into (U,y_1,...,y_m) with y_i(x_1,...,x_m) (with i=1,...,m) smooth functions then the corresponding Jacobi matrix gives rise to an element of Gauge(U) by putting locally g_{ij}(x_1,...,x_m):=dy_i/dx_j.

Hence among gauge transformations there are those which stem from a diffeomorphism hence we get a natural embedding Diff(M) < Gauge(M).

The question is: (after appropriate topologies considered) can we say something about the quotient Gauge(M)/Diff(M) i.e., in what extent is the gauge group "bigger" than the diffeomorhism group of a manifold?

I would expect that the answer splits into a local answer and then a global one (involving the topology of M).

The motivation comes from Kodaira-Spencer deformation theory of complex structures. In this theory two almost complex operators are considered to be equivalent if they differ by a diffeomorphism. However apparently gauge equivalence would be also a natural equivalence relation. Is this beacause simply Kodaira-Spencer theory historically preceded gauge theory?

Thanks!

share|improve this question
8  
The diffeomorphism group is not a subgroup of the gauge group, because a diffeomorphism f induces maps $T_x M \to T_{f(x)} M$, rather than from $T_x M$ to itself. In other words, Df is not a map of bundles over $X$. –  Lucas Culler Sep 20 '10 at 12:05
4  
Something seems a little odd about your map from Diff(M) to Gauge(M). An element of Diff(M) defines an isomorphism T_xM -> T_yM (where x -> y) but an element of Gauge(M) can only define an isomorphism T_xM -> T_xM. –  Loop Space Sep 20 '10 at 12:09
1  
The Diff(M) group can be viewed either in an "active" way carrying point x to y or in a passive way changing the coordinate system about a point (the group of coordinate transformations). I use this second picture. –  E von Tuzzenthaler Sep 20 '10 at 13:48
1  
Even if you work in coordinates, as you do, observe that your map which associates to a diffeomorphism a gauge transformation is not injective. For example the identity and the shift $x\mapsto x+1$ on $\mathbb{R}$ induce the same gauge transformation. –  Michael Bächtold Sep 20 '10 at 17:27
1  
Repeating what Andrew Stacey and Lucas Culler have said in more physics-y language: the Jacobi matrix does not transform as a tensor. So it does not define a section of GL(TM). As a trivial example, let M be the disjoint union of two lines. Pick a coordinate x on one of the lines and a coordinate y on the other one. Then there is a diffeomorphism of the form y(x) = x, x(y) = y. The Jacobi matrix near x=0 is 1 in these coordinates. But under the change of coordinates Y = Y(y), which does not change the x coordinates at all, the Jacobi matrix near x=0 changes to Y'(x). –  Theo Johnson-Freyd Sep 20 '10 at 18:03

1 Answer 1

What you are trying to express, is the following, imho. For the sake of clarity let us split $M$ into two manifolds, $M$, $N$. Consider the 1-jet bundle $\pi_{M\times N}:J^1(M,N)\to M\times N$, which is bundle isomorphic to $L(TM,TN)$. Given smooth $f:M\to N$, we get the 1-jet section $j^1f:M\to J^1(M,N)$ of $\pi_M: J^1(M,N)\to M$ which satisfies $\pi_N\circ j^1f = f:M\to N$.

Now your question is: Given a section $s:M\to J^1(M,N)$ of $\pi_M: J^1(M,N)\to M$, can you recognize when $s=j^1(\pi_N\circ s)$.

Answer: In fact you can. There is a module (over $C^\infty(M)$) of canonical 1-forms (called contact forms or Lepage forms) on $J^1(M,N)$, (edited) locally generated by $dy^j - k^j_i\,dx^i$ in terms of coordinates $(x_i,y^j,k^j_i)$ on $J^1(M,N)$ induced by coordinates $(x^i)$ on $M$ and $(y^j)$ on $N$.

  • We have $s=j^1(\pi_N\circ s)$ if and only if $s^*\omega = 0$ for each contact form. See Wikipedia.

Note that the gauge group $\operatorname{Gau}(M)$ acts from the right on $J^1(M,N)$, and $\operatorname{Gau}(N)$ acts from the left.

share|improve this answer
    
Just noticed that your condition, $\pi_M^* \omega = 0$, on contact forms can't be correct (the arrows point the wrong way). I think what you wanted was to characterize contact forms in a way that easier to check than the defining condition $s^*\omega = 0$ for all $s=j^1 f$. Perhaps the quickest way to do that is to use adapted coordinates on $J^1(M,N)$, say $(x^i,y^j,k^j_i)$. Then, contact forms are all those that are locally generated by the forms $dy^j - k^j_i dx^i$, as you well know of course. –  Igor Khavkine Dec 11 at 23:47
    
@ Igor Khavkine: You are right, I was tired. Thankyou. I changed it. –  Peter Michor Dec 12 at 6:49

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.