I am currently reading through Jost's Riemannian Geometry and Geometric Analysis and am seeking clarification of the statement in the title.
Jost abstractly defines a connection on a vector bundle $D:\Gamma(E)\rightarrow \Gamma(E\otimes T^*M)$ as
$D=d+A$ where $d$ is the exterior derivative and $A=(A^k_j)$ and $A^k_j=\Gamma^k_{ij}dx^i$.
He derives the transformation law for the matrix $A$ as
$A_{\alpha}=\varphi_{\beta\alpha}^{-1}d\varphi_{\beta\alpha}+\varphi_{\beta\alpha}^{-1}A_{\beta}\varphi_{\beta\alpha}$
Where $\varphi_{\beta\alpha}$ are the transition maps associated with the trivialisations of the vector bundle.
Jost then goes on to say
'Thus, $A$ does not transform like a tensor (because of the $\varphi_{\beta\alpha}^{-1}d\varphi_{\beta\alpha}$ term) but the difference of two connection transforms as a tensor'.
I want to get a better understanding of this statement.
I understand how covariant and contravariant tensors transform under a change of coordinates. However, I am struggling to see why the presence of the $\varphi_{\beta\alpha}^{-1}d\varphi_{\beta\alpha}$ term means $A$ does not transform like a tensor.
Finally, in this context of connections on vector bundles, why is it that the difference of two connections transforms as a tensor?
Best Answer
Let me explain here an actually useful characterisation of a "tensor" that is not as oldfashioned as the one of how a "tensor" transforms under change of coordinates. In the process I hope I can clarify why the difference between two connexions is a "tensor".
I shall assume smoothness everywhere.
A connexion, the way it is defined by the Original Poster, $D:\Gamma(E)\to\Gamma(E\otimes T^*M)$ takes smooth sections of the vector bundle $E$ to differential $1$-forms taking values at sections of $E$. One can see this by recognising that $\Gamma(E\otimes T^*M)\cong\Gamma(E)\otimes_{C^\infty(M;\mathbb{R})}\Gamma(T^*M)\cong\mathrm{Hom}_{C^\infty(M;\mathbb{R})}(\mathfrak{X}(M;\mathbb{R});\Gamma(E))$, where I am identifying the $C^\infty(M;\mathbb{R})$-module of sections of the tangent bundle $\Gamma(TM)$ with vector fields of $M$, as derivations on the algebra $C^\infty(M;\mathbb{R})$ ---I do recommend Lee's Introduction to smooth manifolds, or Wald's General Relativity if one is not used to these notions.
That being said, let $s\in\Gamma(E)$ a section of $E$, $X\in\mathfrak{X}(M;\mathbb{R})$ a vector field on $M$ and $f\in C^\infty(M;\mathbb{R})$ a smooth function. By applying the definition of a connexion (as given by the Original Poster), one can readily see that
$$Ds(fX)=fDs(X)$$
and
$$Dfs(X)=X(f)s+fDs(X) \ .$$
Remark: this last property is usually a defining property for a connexion on a vector bundle when it is defined globally, as opposed to the Original Poster's local definition.
Now, if $D$ and $D'$ are two connexions defined on $E$, their difference satisfies
$$(D-D')s(fX)=f(D-D')s(X)$$
and
$$(D-D')fs(X)=f(D-D')s(X) \ .$$ This last equality ---which does not hold for any of the connexions alone--- that characterise their difference as being a "tensor"; their difference is actually a differential $1$-form taking values on the sections of $E$.
A "tensor" is just a mapping that when its arguments are multiplied by functions, it behaves as a linear mapping regarding this $C^\infty(M;\mathbb{R})$-module structure. This is not as precise as I like, but let me show an example with the metric "tensor" (a riemanninan structure $\mathrm{g}$): with vector fields $X,Y,Z\in\mathfrak{X}(M;\mathbb{R})$ and smooth function $f\in C^\infty(M;\mathbb{R})$, the riemannian metric satisfies $\mathrm{g}(X+fY,Z)=\mathrm{g}(X,Z)+f\mathrm{g}(Y,Z)$.
This type of behaviour, with respect to multiplication by functions, guarantees that these "tensors" only depend on what happens at a point. And the way they transform under diffeomorphisms (change of coordinates) can be deduced from that property.