Warning: I am not a differential geometer, so some of the following might not make sense.


Let $w: (T\Omega)^k \to \mathbb{R}$ be a $k$-tensor on $\Omega$, an open subset of $\mathbb{R}^n$.

We can define the "derivative" of this tensor as follows. $Dw:(T\Omega)^{k+1} \to \mathbb{R}$ is a $k+1$-tensor which enjoys the following approximation property

$$ w(p+v_1)(v_2,v_3,...,v_{k+1}) = w(p)(v_2,...,v_{k+1})+Dw(v_1,v_2,...,v_{k+1})+\textrm{Error} $$

Where the error term is order $|v_1||v_2|...|v_{k+1}|$.

If $f$ is a function, $Df$ is the derivative. Here is an example for $w$ a one form:

$w = Pdx+Qdy$

$Dw = \frac{\partial P}{\partial x} dx \otimes dx+ \frac{\partial Q}{\partial x} dx \otimes dy+\frac{\partial P}{\partial y} dy \otimes dx+\frac{\partial Q}{\partial y} dy \otimes dy$

Notice that if you project $D$ of a one form onto the alternating one forms, you get the exterior derivative.

Call a $k$-tensor $w$ closed if there are open sets $U_i$ covering $\Omega$, and $k-1$ tensors $\eta_i$ on $U_i$ with $w = D\eta_i$ on $U_i$. Call a $k$-tensor $w$ exact if there is a global $k-1$ tensor $\eta$ defined on $\Omega$ with $w = D\eta $.

I am interested in the group of closed $k-$tensors mod exact $k-$tensors.

To generalize this beyond subsets of $\mathbb{R}^n$, it seems clear that we would need a connection, since we need to "evaluate" $w$ at ``the same" tangent vectors, but living at different base points. I know enough about connections to know that they allow this to happen, but not much more unfortunately. I do think there should be a group defined analogously to the one I define above for any manifold with a connection.

My question is simply if this group has been studied before, and if so where I should find information about them in the literature. They would possibly encode more information than the de Rham cohomology groups, including data about the geometry of the manifold instead of just the topology.

Also, any information about the basic properties of these groups (really $\mathbb{R}$ -vector spaces) would be appreciated. Are they finite dimensional for reasonable spaces? Do they depend on only the topology?

  • 3
    $\begingroup$ This looks like the global cohomology of $D$-modules. $\endgroup$
    – S. Carnahan
    Jun 17, 2014 at 23:22
  • $\begingroup$ @EvanJenkins: Thanks for the edits! $\endgroup$ Jun 17, 2014 at 23:57
  • $\begingroup$ Maybe I'm misreading this, but why are the exact k-tensors a subgroup of the closed k-tensors? $\endgroup$ Jun 18, 2014 at 16:36
  • $\begingroup$ @AaronBergman They are a subset since they meet the criteria of a closed k-tensor with only one open set $U = \Omega$. They are a subspace since $D$ is a linear map from $k$-tensors to $k+1$ tensors. $\endgroup$ Jun 18, 2014 at 17:44
  • $\begingroup$ I see. I did read too fast -- you're not using closed and exact how I expected. Sorry. $\endgroup$ Jun 18, 2014 at 18:01

2 Answers 2


Let me try to describe a slightly more general situation and hopefully prove my comment above.

Let $V$ be a vector bundle on a manifold $M$ with connection $D : \Gamma(V) \to \Gamma(V \otimes T^*M)$. Let $Z(V)$ be the group of sections $\omega \in \Gamma(V \otimes T^*M)$ where, for an open covering $U_i$ of $M$, there exists local sections $\eta_i \in \Gamma(U_i,V)$ such that $D\eta_i = \omega |_{U_i}$. Let $B(V)$ be the group of sections $\omega \in \Gamma(V \otimes T^*M)$ such that there exists a global section $\nu\in\Gamma(V)$ such that $D\nu = \omega$. Clearly, $B(V) \subset Z(V)$.

I claim that $$ Z(V)/B(V) \cong \check{H}^1(M,\Gamma_f(V,D)) $$ Here $\Gamma_f(V,D)$ is the sheaf of flat sections of $V$ with respect to the connection $D$.

To get the forward, map, let $\omega \in Z(V)$. Then, as above, we can form the $\eta_i \in \Gamma(U_i,V)$. Let $\eta_{ij} = \eta_i - \eta_j \in \Gamma(U_{ij},V)$. We have $D\eta_{ij}=0$ and $\delta\eta_{ij}=0$ where $\delta$ is the Cech differential, so this is a cocyle. Any other representation of $\omega$ subordinate to the same open cover gives rise to a cocycle which differs by a $\delta$-exact term. Finally, $B(V)$ is in the kernel of this map, so we've established the forward direction.

For the reverse direction, do the usual: choose a partition of unity $\sum \gamma_i = 1$ subordinate to the open cover, $U_i$. Given a cocycle $\eta_{ij}$, we can form $$ \rho_i = \sum_k \eta_{ik}\gamma_k \in \Gamma(U_i,V) $$

Let $\zeta_i = D\rho_i = \sum_k \eta_{ik} d\gamma_k$. Then, $$ \zeta_i-\zeta_j = \sum_k (\eta_{ik} - \eta_{jk}) d\gamma_k = \eta_{ij}\, d\left(\sum_k \gamma_k\right) = 0 $$ So, the $\zeta_i$ define an element $\zeta \in Z(V)$. Adding a $\delta$-exact cocycle, $\alpha_{ij} = \alpha_i - \alpha_j$ with $D\alpha_i=0$ adds to $\rho_i$: $$ \sum_k (\alpha_i - \alpha_k)\gamma_k = \alpha_i - \sum_k \alpha_k \gamma_k $$ The first term is annihilated by $D$, and the $D$ of the second gives rise to an element in $B(V)$, so we're done.

To check that the composition works, note that, given $\eta_{ij} = \eta_i - \eta_k$ $$ \eta_i - \rho_i = \eta_i - \sum_k \eta_{ik} \gamma_k = \sum_k \eta_k \gamma_k $$ As above, this is a global section which gives rise to an element in $B(V)$, so the composition is the identity on $Z(V)/B(V)$.

  • $\begingroup$ Thanks! This looks great. I will probably need a couple days to think through this before accepting. I wonder how "mechanical" this interpretation makes the computations. I will again try computing cohomology of the annulus with various connections. Thanks again! $\endgroup$ Jun 19, 2014 at 23:06

I'm not sure I have a direct answer to your question, but here are some thoughts. I would suggest you investigate some of the literature on "flat" connections. It is precisely the flatness condition that allows you to obtain a cochain complex for computing cohomology.

More explicitly, suppose that $E\rightarrow X$ is a vector bundle on a manifold $X$. I will think of a connection as an $\mathbb{R}$-linear map $$\nabla:\Gamma_{E}\rightarrow\Gamma_{E}\otimes_{C^{\infty}(X)}\Omega_X^1,$$ where $\Gamma_E$ is the sheaf of smooth sections of $E$ and $\Omega_X^1$ is the sheaf of $1$-forms on $X$. Note that $\nabla$ naturally gives rise to associated maps $$\nabla:\Gamma_{E}\otimes_{C^{\infty}(X)}\Omega_X^k\rightarrow\Gamma_{E}\otimes_{C^{\infty}(X)}\Omega_X^{k+1}.$$ Flatness is then the condition $$\nabla^2=0,$$ so that you have a cochain complex $$\Gamma_{E}\rightarrow\Gamma_{E}\otimes_{C^{\infty}(X)}\Omega_X^1\rightarrow\Gamma_{E}\otimes_{C^{\infty}(X)}\Omega_X^{2}\rightarrow\ldots.$$ The cohomology of this complex can be interesting. If you take the trivial line bundle over $X$ with the exterior derivative as connection, then you recover the deRham cohomology of $X$. In other cases, you obtain things other than deRham cohomology.

The things you are differentiating are sections of certain vector bundles, and you are differentiating them with a choice of connection. I hope this is useful.

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    $\begingroup$ I don't think this answer the question, since the OP is specifically asking about the cohomology of this cochain complex. $\endgroup$ Jun 17, 2014 at 22:52
  • $\begingroup$ Thanks, this is useful, if only to expose me to some of the language which is out there. $\endgroup$ Jun 17, 2014 at 23:39
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    $\begingroup$ Also, I am not sure that my "cohomology group" arises in a meaningful way from a complex. That is why I phrased it as "locally exact tensors mod globally exact tensors" above. $\endgroup$ Jun 18, 2014 at 0:52

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