2
$\begingroup$

Let $E\to X$ be a complex flat vector bundle, and say $\nabla_0$ and $\nabla_1$ are two flat connections on it. Let $p:X\times[0, 1]\to X$ denote the projection onto the first factor. Is there a way to construct a flat connection $\tilde{\nabla}$ on $p^*E\to X\times[0, 1]$ such that $\tilde{\nabla}|_{X\times 0}=\nabla_0$ and $\tilde{\nabla}|_{X\times 1}=\nabla_1$?

If this is impossible, then is it possible to construct a (not necessarily flat) connection $\nabla'$ on $p^*E\to X\times[0, 1]$ such that $\tilde{\nabla}|_{X\times 0}=\nabla_0$, $\tilde{\nabla}|_{X\times 1}=\nabla_1$ and $\textrm{ch}(\nabla')=$ the rank of $p^*E\to X\times[0, 1]$? Here $\textrm{ch}(\nabla')$ is the Chern character form of $\nabla'$.

Thanks.

Improve this question
$\endgroup$

2 Answers 2

Reset to default
5
$\begingroup$

The second question leads to secondary characteristic classes. Assuming $\nabla'=\tilde\nabla$ is any connection on $X\times[0,1]$ restricting to $\nabla_i$ on $X\times\{i\}$, $i\in\{0,1\}$, its Chern-Simons form is defined as $$\widetilde{\mathrm{ch}}(\tilde\nabla)=\int_0^1\mathrm{ch}(\tilde\nabla)\in\Omega^{\mathrm{odd}}(X;\mathbb C)\;.$$ It is closed if $\nabla_0$, $\nabla_1$ are flat. In this case, its cohomology class $\widetilde{\mathrm{ch}}(\nabla^0,\nabla^1)=[\widetilde{\mathrm{ch}}(\tilde\nabla)]\in H^{\mathrm{odd}}(X;\mathbb C)$ is independent of the choice of $\tilde\nabla$ (subject only to the boundary conditions above). If $\widetilde{\mathrm{ch}}(\nabla^0,\nabla^1)$ does not vanish, the answer to your second question is no. If it does vanish, there could still be other obstructions.

Improve this answer
$\endgroup$
7
  • $\begingroup$ First of all thank you for your answers. Let me change my question a little bit. Let say I have only one flat connection $\nabla$ on $E\to X$. Can I construct a flat connection $\tilde{\nabla}$ on $p^*E\to X\times[0, 1]$ such that $\tilde{\nabla}|_{X\times 0}=\nabla$ and $\tilde{\nabla}|_{X\times t}$ is any arbitrary flat connection for $0<t\leq 1$ and $\tilde{\nabla}|_{X\times 1}$ is not $\nabla$? That means I just want to specify one flat connection. $\endgroup$
    – Ho Man-Ho
    Commented Feb 21, 2016 at 17:14
  • $\begingroup$ Yes, you can, if $\tilde\nabla$ itself is not flat. Easiest example: take $X=S^1$, then all connections are flat. But you can distinguish them by their holonomy, which can be any element in $GL(r,\mathbb C)$, where $r$ is the rank. If the holonomies of $\nabla_0$ and $\nabla_1$ are not conjugate, the flat connections are not isomorphic. Higher dimensional examples can be constructed using paths in the representation variety of $\pi_1(X)$. If $\tilde\nabla$ is flat, then all $\nabla_t$ are isomorphic (use parallel translation in $t$-direction to see this). $\endgroup$
    – Sebastian Goette
    Commented Feb 21, 2016 at 17:33
  • $\begingroup$ It is an instructive exercise to compute the number $\tilde{\mathrm{ch}}(\nabla^0,\nabla^1)[S^1]$ explicitly for two flat connections on $S^1\times\mathbb C\to S^1$. $\endgroup$
    – Sebastian Goette
    Commented Feb 21, 2016 at 17:41
  • $\begingroup$ Thanks, I see. Actually I really need $\tilde{\nabla}$ to be a flat connection. The real question behind this is the Riemann-Roch-Grothendieck theorem for complex flat vector bundle by Bismut-Lott. I know that you also have some work on it. In the imaginary part of this theorem, it is very interesting to me that the real analytic torsion form does not depend on the flat connection on $F\to X$, in contrast to the Bismut-Cheeger eta form which depends on more data. $\endgroup$
    – Ho Man-Ho
    Commented Feb 21, 2016 at 18:13
  • $\begingroup$ I mean if we look at the statement of the RRG for two flat connections, then the difference between $\displaystyle\textrm{Im(CCS)}(H(Z, F|_Z), \nabla_k^{H(Z, F|_Z)})-\int_{X/B}e(TZ, \nabla^{TZ})\cup\textrm{Im(CCS)}(F, \nabla_k^F)$ for $k=0, 1$ is the exterior differential of the same $T$. And I don't know whether it makes sense to say that the path joining two flat connections is not necessarily flat makes RGG theorem to be harder to prove than other secondary index theorems. In the above comment I meant $\tilde{\eta}$ also depends on $\nabla^E$, although it doesn't have to be flat. $\endgroup$
    – Ho Man-Ho
    Commented Feb 21, 2016 at 18:20
5
$\begingroup$

The answer to your first question is of course no. For a counter example, it is enough to consider ordinary closed 1-forms, and you see that a necessary condition for the positive answer would be that all periods of both forms are the same. In the higher rank case, periods are replaced by the monodromy.

More concretely (after Anton Petrunin's comment): Consider the case of $S^1=\mathbb R/2\pi\mathbb Z$ and the trivial line bundle over it with connections $\nabla_0=d$ and $\nabla_1=d+d\varphi.$ Consider also a connection $\tilde\nabla=d+\omega$ on $\mathbb C\to S^1\times [0;1]=:\tilde M$ which restricts to $\nabla_0$ respectively $\nabla_1.$ Then, the curvature of $\tilde\nabla$ is $d\omega$, and by Stokes theorem we get $$2\pi=\int_{\partial\tilde M}\omega=\int_M d\omega.$$ Hence, the curvature of $\tilde\nabla$ does not vanish identically.

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ You answer is correct, but anyone who knows your terminology already knows the answer. Instead I would give an example say $\mathbb C$-bundle over $\mathbb S^1$. $\endgroup$
    – Anton Petrunin
    Commented Feb 21, 2016 at 20:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .