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Assume $\nabla:C^\infty(E) \rightarrow C^\infty(T^*X \otimes E)$ is a covariant derivative and $u$ is an element in the endomorphism bundle $End E$. I'm confused why is the induced connection of $\nabla$ on $End E$ is $\nabla^{End(E)}u=[\nabla,u]$. Any hints?

Proton
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  • If I use $\tilde{\nabla}$ to express for the induced connection on $\mathrm{End},E$, there is a ''fake'' natural way to define it by $(\tilde{\nabla}X)u:=\nabla(Xu)$ where $X$ is a section of endomorphism bundle, $u$ is a section of $E$, however, $\tilde{\nabla}X$ as an endomorphism does not satisfy the $C^\infty-$linearity. – Z. Liu Mar 01 '23 at 10:01

1 Answers1

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If you want an intuitive picture, you can try this way. If $v\in E$, then $u(v)$ is another section of $E$. Now if we compare $u(v)$ at two different points, intuitively (in coordinates): $$ u_y(v_y) \approx u_x(v_x) + \nabla u_x(v_x)\cdot (y-x) $$

But: $$ v_y \approx v_x + \nabla v_x\cdot (y-x) $$

So: $$ u_y(v_y) \approx u_x(v_y - \nabla v_x\cdot (y-x)) + \nabla u_x(v_x)\cdot (y-x) $$

We can rewrite the last expression as: $$ u_y(v_y) \approx u_x(v_y) - u_x(\nabla v_x)\cdot (y-x) + \nabla (u_x(v_x))\cdot (y-x) = u_x(v_y) + [\nabla,u_x]\,v_x\cdot (y-x), $$

which is now linear in $v$ (dipends on $v$ just at the point $x$).

So, the "efficient" connection for $u$ alone is $[\nabla,u]$.

geodude
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    Formally this is just the Leibniz law(http://en.wikipedia.org/wiki/Derivation_(abstract_algebra)) – Xipan Xiao Sep 10 '13 at 15:34
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    Exactly! I usually think of the Leibniz rule as a way of formalizing the notion of "small change". If $xy$ is a varying rectangle, the variation $x,dy + dx,y$ is the "thin" sides that are varying. – geodude Sep 10 '13 at 15:43
  • @geodude Thank you very much! Now I see it. But how to understand $u_x(v_y)$ in the last formula? It seems a little irrational. – Proton Sep 11 '13 at 07:30
  • It is the linear map $u_x$ applied to the vector $u_y$. Yes, they are at different points. That's exactly what a connection does! We are asking how $u$ changes by changing points, but neglecting the intrinsic change of the vector. – geodude Sep 11 '13 at 18:34