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Let $E$ be a vector bundle over a smooth manifold $M$ and $\nabla$ a connection on $E$. If $A$ is an $\text{End}(E)$-valued $1$-form do we have that $$F_{\nabla+A}=F_\nabla+\nabla A+ A \wedge A?$$

I'm a bit confused about the term $\nabla A$ in this expression as $A$ is not really a section of $E$ so I do not know what does this mean.

Tepes
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    A connection on $E$ induces a connection on $\text{End}(E)$ in a natural way. – Ted Shifrin Mar 18 '24 at 18:42
  • Do we need a connection on $T^M \otimes \text{End}(E)$ to be able to take the covariant derivative of $A$ since $A$ is a section of $T^M \otimes \text{End}(E)$? @TedShifrin – Tepes Mar 18 '24 at 21:10
  • Write this all out in terms of a local trivialization so you can understand it concretely. The $1$-form “part” of $A$ is differentiated just using the exterior derivative. Remember that the answer is an $\text{End}(E)$-valued $2$-form. – Ted Shifrin Mar 18 '24 at 21:38

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Write $D=\nabla+A$. Note that $\nabla$ being a connection on $E$ induces in an obvious manner a connection on $\text{End}(E)$; we shall continue to denote this as $\nabla$. Also, $A$ is an $\text{End}(E)$-valued $1$-form, so we can take the exterior covariant derivative $d_{\nabla^{\text{End}(E)}}A$, which again, I’ll henceforth just write as $d_{\nabla}A$.

Assuming you followed along with the various stuff about vector-bundle valued forms I told you about in previous answers, we have: for any $E$-valued $k$-form $\psi$ (we can actually just stick to $k=0$, but it doesn’t simplify the algebra one bit), \begin{align} R_{D}\wedge_{\text{ev}}\psi &=d_{D}^2\psi\\ &=d_{\nabla}(d_D\psi)+A\wedge_{\text{ev}}d_D\psi\\ &=d_{\nabla}(d_{\nabla}\psi+A\wedge_{\text{ev}}\psi)+ A\wedge_{\text{ev}}(d_{\nabla}\psi+A\wedge_{\text{ev}}\psi)\\ &=d_{\nabla}^2\psi+\underbrace{d_{\nabla}(A\wedge_{\text{ev}}\psi)+A\wedge_{\text{ev}}d_{\nabla}\psi}_{=d_{\nabla}A\wedge_{\text{ev}}\psi}+\underbrace{A\wedge_{\text{ev}}(A\wedge_{\text{ev}}\psi)}_{=(A\wedge_{\circ}A)\wedge_{\text{ev}}\psi}\\ &=(R_{\nabla}+d_{\nabla}A+A\wedge_{\circ}A)\wedge_{\text{ev}}\psi. \end{align} Note that in the penultimate step, I used the product rule (keep in mind $A$ is a 1-form so there’s a $(-1)^1$ sign which pops up and hence allows for the simplification), and I used the ‘associativity’ of these wedges. Here, $\wedge_{\text{ev}}$ and $\wedge_{\circ}$ are wedge products relative to evaluations and compositions $\text{End}(E)\oplus E\to E$, $\text{End}(E)\oplus\text{End}(E)\to\text{End}(E)$ respectively.

Since the above equation holds for all $\psi$, it follows that as an equality of $\text{End}(E)$-valued $2$-forms on $M$, we have \begin{align} R_{\nabla+A}&=R_{\nabla}+d_{\nabla}A+A\wedge_{\circ}A. \end{align}

peek-a-boo
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  • To be completely honest, I'm still not entirely comfortable with working with this $\wedge_{\text{ev}}$ and it seems that my difficulties revolve around that. I started off initially started with

    $$ \begin{align} d^2_{\nabla+A}\varphi&= (\nabla+A)(\nabla \varphi+A\varphi) \ &= \nabla(\nabla \varphi+A\varphi) + A(\nabla \varphi+A\varphi) \ &= d_\nabla(d_{\nabla+A}\varphi) + A(\nabla \varphi+A\varphi) \end{align} $$

    which looks similar to what you wrote, but I don't know how to rephrase the term $A(\nabla \varphi+A\varphi)$ using the evaluation pairing. @peek-a-boo

     – Tepes Mar 19 '24 at 10:56
  • If I blindly go distributing things over I end up with

    $$ \begin{align} d^2_{\nabla+A}\varphi&= (\nabla+A)(\nabla \varphi+A\varphi) \ &= \nabla\nabla \varphi+\nabla A\varphi + A\nabla \varphi + A(A\varphi) \ &= d_\nabla^2\varphi+ \nabla A\varphi + A\nabla \varphi + A(A\varphi)\end{align} $$

    where the term $\nabla A\varphi + A\nabla \varphi$ looks a lot like the product rule and $A(A\varphi)$ like $A \wedge A\varphi$, but I can't justify any of these operations. @peek-a-boo

     – Tepes Mar 19 '24 at 11:51
  • If you want to blindly go over and distribute things then re-read my answer and ignore all the subscripts on $\wedge$. It’s just basic additivity and one usage of the product rule, and associativity of the wedge. – peek-a-boo Mar 19 '24 at 18:33
  • Once you do that I really suggest going over the links I provided previously. They show that for 95% of the time, these various bilinear-pairing wedges $\wedge_{\beta}$: they’re bilinear, “associative”, “anti-commutative”, and when interacting with $d_{\nabla}$, satisfy a product rule. So it really is almost like the usual wedge. – peek-a-boo Mar 19 '24 at 18:36
  • $R_{D}\wedge_{\text{ev}}\psi$ should be particularly simple, this acts on vector fields $X$ and $Y$ by $$R_{D}\wedge_{\text{ev}}\psi(X,Y) = R_D(X,Y)\psi?$$ However the term $A\wedge_{\text{ev}}(d_{\nabla}\psi+A\wedge_{\text{ev}}\psi)$ is already a bit more involved. What kind of an object is this and how does it act? @peek-a-boo – Tepes Mar 19 '24 at 18:50
  • If $k=0$ then it is an $E$-valued $2$-form on $M$. I could tell you how it acts on a pair of vectors, but can you hazard a guess? Also, note that I’ve already given you the definition before here; section 2 of my answer in particular. – peek-a-boo Mar 19 '24 at 18:58
  • Given the definition in section 2, we have for $A \wedge_{\text{ev}} d_\nabla \psi$ that

    $$ A \wedge_{\text{ev}} d_\nabla \psi(X_1,X_2)=\sum_{\sigma \in S_2}\text{sgn}(\sigma)\text{ev}(A(X_{\sigma(1)}), d_\nabla\psi(X_{\sigma(2)})), $$

    which gives $A(X_1)d_\nabla\psi(X_2)-A(X_2)d_\nabla\psi(X_1)$ and the terms $A(X_i)d_\nabla\psi(X_j)$ are matrix-vector products, i.e. vectors consisting of sections of $E$? @peek-a-boo

     – Tepes Mar 19 '24 at 20:02
  • Let us continue this discussion in chat. – peek-a-boo Mar 19 '24 at 21:12