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Let $M$ be an $m$ dimensional differentiable manifold. Define $\text{Gauge}(M) := \text{C}^∞(M, \text{Aut}(TM))$ to be the group of all (smooth) fiberwise linear transformations of the tangent bundle. This is the natural gauge group of the manifold. If $\left(U, x_1,…,x_m\right)$ is a local coordinate system with induced frame on $TU$ then an element of $\text{Gauge}(U)$ looks like an invertible matrix $g_{ij}\left(x_1,…,x_m\right)$ (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 $\left(U,x_1,…,x_m\right)$ into $\left(U,y_1,…,y_m\right)$ with $y_i\left(x_1,…,x_m\right)$ (with $i=1,…,m$) smooth functions then the corresponding Jacobi matrix gives rise to an element of $\text{Gauge}(U)$ by putting locally $$g_{ij}\left(x_1,…,x_m\right) := \frac{dy_i}{dx_j}.$$

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

The question is: (after appropriate topologies considered) can we say something about the quotient $\text{Gauge}(M)/\text{Diff}(M)$ 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 because simply Kodaira-Spencer theory historically preceded gauge theory?

Thanks!

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    $\begingroup$ 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 $X$. $\endgroup$
    – Lucas Culler
    Commented Sep 20, 2010 at 12:05
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    $\begingroup$ 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. $\endgroup$
    – Andrew Stacey
    Commented Sep 20, 2010 at 12:09
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    $\begingroup$ 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. $\endgroup$
    – E von Tuzzenthaler
    Commented Sep 20, 2010 at 13:48
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    $\begingroup$ 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. $\endgroup$
    – Michael Bächtold
    Commented Sep 20, 2010 at 17:27
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    $\begingroup$ 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). $\endgroup$
    – Theo Johnson-Freyd
    Commented Sep 20, 2010 at 18:03

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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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    $\begingroup$ 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. $\endgroup$
    – Igor Khavkine
    Commented Dec 11, 2014 at 23:47
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    $\begingroup$ @ Igor Khavkine: You are right, I was tired. Thankyou. I changed it. $\endgroup$
    – Peter Michor
    Commented Dec 12, 2014 at 6:49

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