# The gauge group versus the diffeomorphism group of a manifold

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!

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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
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
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
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
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

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.

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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