3
$\begingroup$

I'm studying differential geometry, and when I began to study connections, the first definition that I found was the following:

Definition: Let M a differentiable manifold. A connection on M is a transformation $\nabla: D(M)\times D(M)\rightarrow D(M)$, where $D(M)$ is the set of differentiable vector fields on $M$, which satisfices:

a)$\nabla_{fX_1+X_2}Y=f\nabla_{X_1}Y+\nabla_{X_2}Y$ with $f\in C^{\infty}(M)$ and $X_1,X_2, Y\in D(M)$

b)$\nabla_X(\lambda Y_1+Y_2)=\lambda\nabla_X Y_1+\nabla_X Y_2$, with $\lambda\in\mathbb{R}$ and $X,Y_1,Y_2\in D(M)$

c)$\nabla_X(fY)=f\nabla_X Y+X(f)Y$, where $f\in C^{\infty}(M)$ and $X,Y\in D(M)$

I had no problem with this definition, but later the book says that we can reinterpret the previous definition, and we can say that a connection is actually a transformation $\nabla:\Gamma(TM)\rightarrow\Gamma(T^*M\otimes TM)$. Explicitly, if $Y\in\Gamma(TM)$, then $\nabla Y$ will be the element of $\Gamma(T^*M\otimes TM)$ which satisfices: $$\nabla Y(X,\theta)=\theta(\nabla_X Y)$$. So, with this, we can generalize the definition of a connection, but this time in a vector bundle, as follows:

Definition: Let $\xi=(E,\pi)$ a differentiable vector bundle over a differentiable manifold $M$. A connection on $\xi$ is a transformation: $$\nabla:\Gamma(E)\rightarrow\Gamma(T^*M\otimes E)$$ with the following properties:

a)$\nabla(s)(fX+X',\theta)=f\nabla(s)(X,\theta)+\nabla(s)(X',\theta)$.

b)$\nabla(\lambda s+s')=\lambda\nabla s+\nabla s'$

c)$\nabla(fs)=f\nabla s+df\otimes s$

for all $s,s´\in\Gamma(E)$, $X,X'\in\Gamma(TM)$, $\theta\in\Gamma(E^*)$, $f\in C^{\infty}(M)$ and $\lambda\in\mathbb{R}$.

My problem is that I cannot find the way of joining both definitions. If I take, as particular case, $E=TM$ in the second definition, I don't see why the connection defined in this way is the same (or is connected) with the first one . If the second definition is more general, then it should to reduce to the first one when I take $E=TM$. My little book doesn't explain more, and begins to construct the connection one-forms $\omega_{ij}$. If it is important, my book is "Geometría Riemanniana" of Héctor Sánchez Morgado and Oscar Palmas Velasco.

$\endgroup$

2 Answers 2

3
$\begingroup$

The thing here is that, to establish the second definition, and write a connection on a manifold $M$ as a transformation $\nabla:\Gamma(TM)\rightarrow\nabla(T^*M\otimes TM)$, it is neccesary to use the isomorphism between $\text{Hom}(TM,TM)$ and $T^*M\otimes TM$ in the following way:

Let $\nabla:\mathfrak{X}(M)\times\mathfrak{X}(M)\rightarrow\mathfrak{X}(M)$ a connection on $M$ and $Y\in\mathfrak{X}(M)$. If $p\in M$, we define:

$$\nabla Y(p):T_pM\rightarrow T_pM$$

as $(\nabla Y(p))(v):=(\nabla_X Y)(p)$, with $v\in T_pM$ and $X\in\mathfrak{X}(M)$ any differentiable vector field on $M$ such that $X(p)=v$. So, in particular, $\nabla Y(p)$ will be a linear function, and then:

$$\nabla Y(p)\in\text{Hom}(T_pM,T_pM)$$

which means that $\nabla Y\in\Gamma(\text{Hom}(TM,TM))$. But the bundle $\text{Hom}(TM,TM)$ is isomorphic to $T^*M\otimes TM$. This means that we can consider that:

$$\nabla Y\in\Gamma(T^*M\otimes TM)$$

Based on the above, we can reinterpret a connection on $M$ as a transformation:

$$\nabla:\Gamma(TM)\rightarrow\Gamma(T^*M\otimes TM)$$.

$\endgroup$
1
$\begingroup$

If $E\to M$ is a vector bundle, a Koszul connection for $E$ is a map $\nabla:\mathfrak{X}(M)\times \Gamma(E)\to \Gamma(E)$, taking $(X,\psi)$ to a section $\nabla_X\psi$, such that $\nabla$ is $\mathcal{C}^\infty(M)$-linear in the first entry, $\Bbb R$-linear in the second entry, and satisfies a Leibniz rule in the second entry: $$\nabla_X(f\psi)=X(f)\psi+f\nabla_X\psi.$$The value of $(\nabla_X\psi)_x$, for $x\in M$, depends only on the value $X_x$ and on the values of $\psi$ in a neighborhood of $x$. Fixed coordinates $(x^j)$ for $M$ and a local trivialization $(e_a)$ for $E$, we have "Christoffel symbols" $$\nabla_{\partial_j}e_a =\sum_b \Gamma_{ja}^be_b,$$and so on. Alternatively, one may write $$\nabla_Xe_b = \sum_a \omega^a_{~b}(X)e_a,$$for local $1$-forms $\omega^a_{~b}$. It pays off and helps to avoid confusion to use different alphabets to index things living in different worlds, by the way. Also note that while I'm still writing summation signs for pedagogical reasons, I'm still respecting the index balance required by Einstein's convention (this helps detecting errors in expressions and keeps things natural).

$\endgroup$
1
  • $\begingroup$ Thank you, but the thing you're mentioning doesn't answer my question. $\endgroup$ Commented Mar 26, 2020 at 0:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .