# 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. –  Andrew Stacey 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
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