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I am trying to connect two different notions of connection I learned:

I first learned the notion of connections in Demailly's "Complex Analytic and Differential Geometry", where in the section called "Linear Connections", he defined a connection $D$ on a bundle $E$ as

a linear differential operator of order 1 acting on $\mathcal{C}_*^\infty(M,E)$ and satisfying the following properties: $$\text{(i)} D:\mathcal{C}_q^\infty(M,E)\to \mathcal{C}_{q+1}^\infty(M,E)$$ $$\text{(ii)} D(f\wedge s)=df\wedge s+(-1)^q f\wedge Ds$$ for any $f\in \mathcal{C}_p^\infty(M,K)$ and $s\in \mathcal{C}_q^\infty(M,E)$.

Here he appears to use $\mathcal{C}_q^\infty(M,E)$ to denote the bundle $\bigwedge^q T^*M\otimes E$

At the start of "Elliptic operators, topology and asymptotic methods", the author defines a connection on a vector bundle $V$ as

a linear map $$\nabla: C^\infty(TM)\otimes C^\infty(V)\to C^\infty(V)$$ assigning to a vector field $X$ and a section $Y$ of $V$ a new vector field $\nabla_X Y$ such that, for any smooth function $f$ on $M$, $$\text{(i) }\nabla_{fX} Y=f\nabla_X Y$$ $$\text{(ii) }\nabla_X(fY)=f\nabla_X Y+(X.f)Y,$$ where $X.f$ denotes the Lie derivatives of $f$ along $X$.

Are these two notions of connection the same? Would $\nabla_X\nabla_Y(s)$ be the same as $D^2(s)$ evaluated on $(X,Y)$? (I am not sure how to evaluate a section of $\bigwedge^2 T^*M\otimes E$ on a pair of vector fields, so this is my best guess of how could these two be equivalent.)

kid111
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1 Answers1

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The first definition sounds much more like the exterior covariant derivative associated to a linear connection (as in the second definition); another common notation for it is $d^{\nabla}$ or $d_{\nabla}$. Also, in the second definition, you write

assigning to a vector field $X$ and a section $Y$ of $V$ a vector field $\nabla_XY$...

the second instance of vector field (emphasis mine) is a typo, it should say "a section $\nabla_XY$ of $V$".

Also, in the first definition, $D^2s$, the second covariant exterior differential of a smooth section $E$-valued $q$ form $s$ on $M$, equals the 'wedge' of the curvature, $R$, of the connection (which is a smooth $\text{End}(E)$-valued $2$-form on $M$, i.e $R\in \mathcal{C}^{\infty}_2(M,\text{End}(E))$) and $s$: \begin{align} D^2s&=R\wedge_{\epsilon}s. \end{align} Here, $\wedge_{\epsilon}$ denotes the wedge product with respect to evaluation of endomorphisms on vectors: see Ivo Terek's answer here for details. Its value on a pair of vector fields $X,Y$ is \begin{align} (D^2s)(X,Y)&=\nabla_X\nabla_Ys-\nabla_Y\nabla_Xs-\nabla_{[X,Y]}s. \end{align}

As far as terminology is concerned, I find this usage confusing, but the operators are essentially the same when acting on sections of $E$, i.e $D:\mathcal{C}^{\infty}_0(M,E)\to \mathcal{C}^{\infty}_1(M,E)$ and $\nabla:C^{\infty}(TM)\otimes C^{\infty}(E)\to C^{\infty}(E)$ are related as: \begin{align} (Ds)(X)&=\nabla_Xs. \end{align} Or, $Ds=\nabla s$, where it is understood that there is an open slot to be filled in by a vector field $(Ds)(\cdot)=\nabla_{(\cdot)}s$; this open slot for a vector field is what gives the 1-form nature.

peek-a-boo
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  • I got the part when you say the operators are the same when acting on sections of $E$, however, I wonder whether if for some (or all perhaps?) of $D:\mathcal{C}^\infty_{n}(M,E)\to \mathcal{C}^\infty_{n+1}(M,E)$, there is a corresponding connection, in the sense of the second definition, on the bundle $\bigwedge^n T^*M\otimes E$ that induce $D$? – kid111 Jun 10 '22 at 16:19
  • @kid111 the exterior covariant derivative $D=d_{\nabla}$ (starting from a connection $\nabla$ on $E$ in the sense of the second definition) maps for all $q\geq 0$, sections of $\left(\bigwedge^q(T^M)\right)\otimes E$ to sections of $\left(\bigwedge^{q+1}(T^M)\right)\otimes E$. However, if you have a covariant derivative $\tilde{\nabla}$ on $\left(\bigwedge^q(T^M)\right)\otimes E$ (in terms of the second definition), then it maps sections of this bundle to sections of $T^M\otimes \left(\left(\bigwedge^q(T^*M)\right)\otimes E\right)$. It is only for $q=0$ that these spaces coincide. – peek-a-boo Jun 10 '22 at 17:03
  • So, instead, here's what I prefer to think of. A connection, is most conveniently defined using Ehresmann's definition; I actually asked a question about it before; I think the picture I drew might be helpful. So, now that we have a direct sum decomposition $TE=VE\oplus HE$, we can construct several types of differential operators. The first is $\nabla$, which takes sections of $E$ to sections of $T^*M\otimes E$. – peek-a-boo Jun 10 '22 at 17:08
  • The definition of operator $\nabla$ is that if you have a section $s:M\to E$, then the tangent map is $Ts:TM\to TE$. Using the direct sum decomposition, you get a map $P_{VE}\circ Ts:TM\to VE$, and finally since the vertical bundle can be identified with $E$, you get a mapping $\nabla s:TM\to E$. So, this gives us a way of differentiate $E$-valued objects. Now, as with the usual exterior derivative, we can 'antisymmetrize' $\nabla$ to get a collection of maps $d_{\nabla}$ which map $\mathcal{C}^{\infty}q(M,E)\to \mathcal{C}^{\infty}{q+1}(M,E)$, and this is the exterior covariant derivative. – peek-a-boo Jun 10 '22 at 17:11