Author: Eiko

Tags: differential geometry, connection, curvature, differential algebra

Time: 2024-10-17 14:13:27 - 2024-10-20 17:50:00 (UTC)

Connection, Covariant Derivative

Given a vector field $v$, ${\mathrm{\nabla }}_v$ is an operator defined on the following spaces

$${\mathrm{\nabla }}_v:TM\to TM,{\mathrm{\nabla }}_v:{\mathcal{O}}_M\to {\mathcal{O}}_M$$

Which are completely determined by the two requirements ${\mathrm{\nabla }}_{v}f=\frac{\partial f}{\partial v}$ and ${\mathrm{\nabla }}_{e_j}{e}_i={\mathrm{\Gamma }}_{ij}^k{e}_k$.

By extending tensorially (by multiplicative law), and extending dually (by the contravariant Hom functor), we can define the covariant derivative on any tensors of $TM$ and $T^{\ast }M$, as well as $\mathrm{End}(TM)=TM\otimes T^{\ast }M$.

Remark

There might be many choices of connections / covariant derivatives, on a Riemannian manifold there is a unique and canonical choice called Levi-Civita connection that it is torsion-free and metric-compatible.

Eta reduction and currying

If we reduce the $v$ inside $\mathrm{\nabla }$, then $\mathrm{\nabla }$ will be an operator that attach a covector tensor term to any tensor bundle $V$

$$\mathrm{\nabla }:V\to T^{\ast }M\otimes V.$$

In terms of coordinates, for example for $V=TM$ we have the formula

$$\mathrm{\nabla }:TM\to TM\otimes T^{\ast }M,\mathrm{\nabla }u=\sum _j(u_{;j}^i{e}_i+u^i{\mathrm{\Gamma }}_{ij}^k{e}_k)\otimes d{x}^j.$$

• Notice that on $TM\otimes T^{\ast }M=\mathrm{End}(TM)$ there is a natural trace map, which gives the map $TM\stackrel{\mathrm{\nabla }}{\to }\mathrm{End}(TM)\stackrel{\mathrm{tr}}{\to }{\mathcal{O}}_M$

  $$u\mapsto u_{;j}^j+u^i{\mathrm{\Gamma }}_{ij}^j$$

  and this is the divergence operator on vector fields!

• The map $\mathrm{\nabla }:TM\to \mathrm{End}(TM)$ means that covariant derivative is making $TM$ a $TM$-module! ($TM$ is not a ring though).

• We can try to extend the module structure to an actual ring, the tensor ring $T^{\bullet }(TM)=\underset{n}{⨁}(TM{)}^{\otimes n}$, so now $TM$ is a $T^{\bullet }(TM)$-module. (Curious: If this module does not give up things like curvature, why does $D$-modules have to give up them and work on flat connections only? We will see.)

Curvature Tensor

Second Covariant Derivative

How to define the second covariant derivative ${\mathrm{\nabla }}^2$? There are two candidates:

• The twice covariant derivative ${\mathrm{\nabla }}_u\circ {\mathrm{\nabla }}_v$ can be reduced as

  $${\mathrm{\nabla }}^2:V\to T^{\ast }M\otimes T^{\ast }M\otimes V.$$

  This one is just ${\mathrm{\nabla }}_{Y,X}^2={\mathrm{\nabla }}_Y\circ {\mathrm{\nabla }}_X$.

• Define it as $V\to T^{\ast }M\otimes V\to T^{\ast }M\otimes (T^{\ast }M\otimes V)$ where in the second arrow we take derivative for the whole $T^{\ast }M\otimes V$.

  $${\mathrm{\nabla }}^2:V\to T^{\ast }M\otimes T^{\ast }M\otimes V.$$

  This is a bit different, in the first arrow it maps $u\mapsto {\mathrm{\nabla }}_{◻}u$, and in the second arrow we need to differentiate on both $◻$ and $u$. Differentiation on the hole $◻$ need to take into account a negative sign and then compose ${\mathrm{\nabla }}_Y$ on the right, because of dualizing. We get

  $${\mathrm{\nabla }}_Y({\mathrm{\nabla }}_{◻}u)=({\mathrm{\nabla }}_Y\circ {\mathrm{\nabla }}_{◻})u-{\mathrm{\nabla }}_{{\mathrm{\nabla }}_Y◻}u,$$

  therefore

  $${\mathrm{\nabla }}_{Y,X}^2={\mathrm{\nabla }}_Y{\mathrm{\nabla }}_X-{\mathrm{\nabla }}_{{\mathrm{\nabla }}_{Y}X}.$$

It seems that people prefer the second definition.

Curvature Tensor

The Riemann curvature tensor is defined as

$$R(X,Y)=[{\mathrm{\nabla }}_X,{\mathrm{\nabla }}_Y]-{\mathrm{\nabla }}_{[X,Y]}={\mathrm{\nabla }}_{X,Y}^2-{\mathrm{\nabla }}_{Y,X}^2,$$

which measures the difference of taking the derivative in two different orders. The second derivative actually exhibits this point.

Connections On General Bundles

Given any vector bundle $V$ on $M$ one can similarly define connection as a map

$$\mathrm{\nabla }:V\to {\mathrm{\Omega }}_M^1\otimes V.$$

Starting here we will write ${\mathrm{\Omega }}_M^1$ instead of $T^{\ast }M$. Imagine them as the same thing!

Taking iterated covariant derivatives should give us a sequence of maps

$$V\to {\mathrm{\Omega }}^1\otimes V\to {\mathrm{\Omega }}^1\otimes ({\mathrm{\Omega }}^1\otimes V)\to {\mathrm{\Omega }}^1\otimes ({\mathrm{\Omega }}^1\otimes ({\mathrm{\Omega }}^1\otimes V))\to \cdots$$

For example, the second derivative is given by ${\mathrm{\nabla }}_{X,Y}^2={\mathrm{\nabla }}_X{\mathrm{\nabla }}_Y-{\mathrm{\nabla }}_{{\mathrm{\nabla }}_{X}Y}$, and

$${\mathrm{\nabla }}_{X,Y,Z}^3={\mathrm{\nabla }}_X{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_Z-{\mathrm{\nabla }}_X{\mathrm{\nabla }}_{{\mathrm{\nabla }}_{Y}Z}-{\mathrm{\nabla }}_{{\mathrm{\nabla }}_{X}Y}{\mathrm{\nabla }}_Z-{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_{{\mathrm{\nabla }}_{X}Z}-{\mathrm{\nabla }}_{{\mathrm{\nabla }}_{{\mathrm{\nabla }}_{X}Y}Z}+{\mathrm{\nabla }}_{{\mathrm{\nabla }}_X{\mathrm{\nabla }}_{Y}Z}.$$

Collapse Into Wedge Forms

In order to relate the above construction induced from connection to the de-Rham complex and cohomology theories, we can define another complex similar to the above sequence, with spaces replaced by ${\mathrm{\Omega }}_M^k\otimes V$ instead of $({\mathrm{\Omega }}^1{)}^k\otimes V$.

To do this, we recall the following constructions on vector spaces

• ${\wedge }^{k}V\stackrel{{\iota }_k}{\hookrightarrow }{V}^{\otimes k}$, the inclusion of antisymmetric tensors into symmetric tensors given by

  $${\iota }_k(v_1\wedge \cdots \wedge v_k)=\sum _{\sigma \in S_k}\mathrm{sgn}(\sigma )v_{\sigma (1)}\otimes \cdots \otimes v_{\sigma (k)}.$$

• $V^{\otimes k}\stackrel{{\alpha }_k}{\to }{\wedge }^{k}V$, the projection of symmetric tensors into antisymmetric tensors given by

  $${\alpha }_k(v_1\otimes \cdots \otimes v_k)=\frac{1}{k!}{v}_1\wedge \cdots \wedge v_k.$$

We have that $\alpha \circ \iota =\mathrm{id}$, so the sequencd $0\to {\mathrm{Sym}}^{k}V\stackrel{}{\to }{V}^{\otimes k}\stackrel{\alpha }{\to }{\wedge }^{k}V\to 0$ is split exact. Therefore we define the following complex as

Clearly ${\alpha }_0=\mathrm{id}$, ${\alpha }_1=\mathrm{id}$.

• On the first line ${\mathrm{\Omega }}^1\otimes V\to {\mathrm{\Omega }}^1\otimes ({\mathrm{\Omega }}^1\otimes V)$, things are defined as

$${\mathrm{\nabla }}_{◻}(\omega \otimes v)={\mathrm{\nabla }}_{◻}\omega \otimes v+\omega \otimes {\mathrm{\nabla }}_{◻}v.$$

$$({\mathrm{\nabla }}_{◻}(\omega \otimes v))(X,Y)=({\mathrm{\nabla }}_X\omega )(Y)v+\omega (Y){\mathrm{\nabla }}_{X}v.$$

• On the second line ${\mathrm{\Omega }}^1\otimes V\to {\mathrm{\Omega }}^2\otimes V$, things are induced from the first line, let’s observe what is ${\iota }_2\circ {\alpha }_2({\mathrm{\nabla }}_{◻}(\omega \otimes v))$:

  $$\begin{array}{rl}2{\iota }_2\circ {\alpha }_2({\mathrm{\nabla }}_{◻}(\omega \otimes v))(X,Y)& =({\mathrm{\nabla }}_X\omega )(Y)v-({\mathrm{\nabla }}_Y\omega )(X)v+\omega (Y){\mathrm{\nabla }}_{X}v-\omega (X){\mathrm{\nabla }}_{Y}v\end{array}$$

• compare it with the definition that $\mathrm{\nabla }(\omega \otimes v)=d\omega \otimes v-\omega \wedge \mathrm{\nabla }v$, which expands to

  • The first term is $$(d\omega \otimes v)(X,Y)=[({\mathrm{\nabla }}_X\omega )(Y)-({\mathrm{\nabla }}_Y\omega )(X)+\omega ({\mathrm{\nabla }}_{X}Y-{\mathrm{\nabla }}_{Y}X-[X,Y])]v,$$

    where $d\omega$ is computed as

    $$\begin{array}{rl}(d\omega )(X,Y)& =X(\omega (Y))-Y(\omega (X))-\omega ([X,Y])\\ & ={\mathrm{\nabla }}_X⟨\omega ,Y⟩-{\mathrm{\nabla }}_Y⟨\omega ,X⟩-⟨\omega ,[X,Y]⟩\\ & =⟨{\mathrm{\nabla }}_X\omega ,Y⟩+⟨\omega ,{\mathrm{\nabla }}_{X}Y⟩-⟨{\mathrm{\nabla }}_Y\omega ,X⟩-⟨\omega ,{\mathrm{\nabla }}_{Y}X⟩-⟨\omega ,[X,Y]⟩\\ & =({\mathrm{\nabla }}_X\omega )(Y)-({\mathrm{\nabla }}_Y\omega )(X)+\omega ({\mathrm{\nabla }}_{X}Y-{\mathrm{\nabla }}_{Y}X-[X,Y]).\end{array}$$

  • The second term is $$(\omega \wedge \mathrm{\nabla }v)(X,Y)=\omega (X){\mathrm{\nabla }}_{Y}v-\omega (Y){\mathrm{\nabla }}_{X}v,$$

  So this definition would give us

  $$({\mathrm{\nabla }}_X\omega )(Y)v-({\mathrm{\nabla }}_Y\omega )(X)v+\omega ({\mathrm{\nabla }}_{X}Y-{\mathrm{\nabla }}_{Y}X-[X,Y])v-\omega (X){\mathrm{\nabla }}_{Y}v+\omega (Y){\mathrm{\nabla }}_{X}v$$

  If we add an assumption that our connection be torsion free (which states ${\mathrm{\nabla }}_{X}Y-{\mathrm{\nabla }}_{Y}X=[X,Y]$), then the this definition coincides with the definition induced from the first row (only differ by a constant factor $k!$).

Remark:

• ${\alpha }_k$ and ${\iota }_k$ are natural transformations between polynomial functors ${\otimes }^k$ and ${\wedge }^k$.

Algebraic Connections

The above construction interprets the complex ${\mathrm{\Omega }}^i\otimes V$ as a subcomplex of the connection in geometry $({\mathrm{\Omega }}^1{)}^i\otimes V$, where things are all anti-symmetric.

With the above correspondence and intuition, we can now define a purely algebraically defined connection on the complex ${\mathrm{\Omega }}^{\bullet }\otimes V$, without referring to the first row.

The rule of connection ${\mathrm{\nabla }}^{(p)}:{\mathrm{\Omega }}^p\otimes V\to {\mathrm{\Omega }}^{p+1}\otimes V$ is defined as

$$\mathrm{\nabla }(\omega \otimes v)=d\omega \otimes v+(-1{)}^p\omega \wedge \mathrm{\nabla }v,\omega \in {\mathrm{\Omega }}^p,v\in V.$$

Curvature comes in automatically

Because we are working in anti-symmetric tensors, the curvature tensor is always here as long as you try to differentiate twice.

$$\begin{array}{rl}{\mathrm{\nabla }}^2(\omega \otimes v)& =\mathrm{\nabla }(d\omega \otimes v+(-1{)}^p\omega \wedge \mathrm{\nabla }v)\\ & =d(d\omega )\otimes v+(-1{)}^{p+1}d\omega \wedge \mathrm{\nabla }v+(-1{)}^{p}d\omega \wedge \mathrm{\nabla }v+(-1{)}^{2p}\omega \wedge {\mathrm{\nabla }}^{2}v\\ & =\omega \wedge {\mathrm{\nabla }}^{2}v\\ & =\omega \wedge (X,Y\mapsto {\mathrm{\nabla }}_{X,Y}^{\prime 2}v-{\mathrm{\nabla }}_{Y,X}^{\prime 2}v)\\ & =\omega \wedge (X,Y\mapsto R(X,Y))v\\ & =\omega \wedge Rv\end{array}$$

here ${\mathrm{\nabla }}^{\prime }$ means the original connection (the one in the first row, without implicit anti-symmetrization).

Therefore, the connection $\mathrm{\nabla }$ is called flat or integrable if ${\mathrm{\nabla }}^2=0$, i.e. the curvature tensor vanishes. When that happens, we have a cool complex $({\mathrm{\Omega }}^{\bullet }\otimes V,\mathrm{\nabla })$ that can be used to define algebraic de-Rham cohomology.