# Regarding the definition of connections (covariant derivatives)

A connection can be thought of as a map $\Gamma(TM) \times \Gamma(TM) \to \Gamma(TM)$, so $\nabla_X Y$ is section of the tangent bundle. But it can also be thought of as a map $\Gamma(TM) \to \Gamma(T^*M \otimes TM)$, but why is that? I don't see how.

I thought maybe one keeps $X$ fixed in $\nabla_X Y$ but this doesn't give me a section of $T^*M \otimes TM$. Can someone explain it, please?

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## 2 Answers

A section of $TM \otimes T^\ast M$ is just a $TM$-valued $1$-form. In other words, a section $\sigma \in \Gamma(TM \otimes T^\ast M)$ can be considered as a linear map $$\tilde{\sigma}: \Gamma(TM) \longrightarrow \Gamma(TM),$$ $$\tilde{\sigma}(p, v) = \sigma_p(v).$$

For a concrete example, consider the constant section $(1,2) \otimes dx \in \Gamma(T\Bbb R^2 \otimes T^\ast \Bbb R^2)$. Then the associated linear map $\Gamma(T\Bbb R^2) \longrightarrow \Gamma(T\Bbb R^2)$ in this case is $$(p, v) \mapsto ((1,2) \otimes dx)_p(v) = (1,2) \cdot dx_p(v) = (p,(v_1,2v_1)),$$ where $v = (v_1, v_2)$.

If you want to view a connection as a map $\nabla: \Gamma(TM) \longrightarrow \Gamma(TM \otimes T^\ast M)$ as opposed to a map $\nabla_X : \Gamma(TM) \longrightarrow \Gamma(TM)$ for some vector field $X$, then you should imagine it as the map $$Y \mapsto (X \mapsto \nabla_X Y).$$ In other words, you instead think of $X$ as the argument of the associated linear map $$\nabla_\bullet Y : \Gamma(TM) \longrightarrow \Gamma(TM).$$ For each choice of $X$ you then get a section $\nabla_X Y \in \Gamma(TM)$, so you should be able to see that $\nabla_\bullet Y$ is a section of $TM \otimes T^\ast M$.

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Thanks for the answer. Is the section $(1,2)\otimes dx$ the same as $1\partial_{x_1} \otimes 2\partial_{x_2}\otimes dx^1 \otimes dx^2$? I have just started with this stuff and am used to thinking about it in terms of basis.. – soup Nov 26 '12 at 21:17
I was using coordinates $(x,y)$ instead of $(x^1, x^2)$ here. What I mean is $(1 \cdot \partial_x + 2 \cdot \partial_y) \otimes (1 \cdot dx + 0 \cdot dy) = (\partial_x + 2\partial_y) \otimes dx$. – Henry T. Horton Nov 26 '12 at 21:23

It does, if you identify $T^*M \otimes TM$ with $\text{Hom}(TM, TM)$. I'll describe the identification in the linear algebra setting between $V^* \otimes W$ and $Hom(V,W)$. Pick a basis $(v_1, ..., v_n)$ for $V$, take $(\phi^1, ..., \phi^n)$ to be the dual basis of $(v_i)$ for $V^*$ and take $(w_1, ..., w_m)$ to be a basis of $W$. Elements of $V^* \otimes W$ are of the form $\sum a_i^j \phi^i \otimes w_j$. How can we interpret them as maps from $V$ to $W$? Given a vector $v \in V$, the action of this element on $v$ is $$\left( \sum a_i^j \phi^i \otimes w_j \right)\left( v \right) = \sum a^j_i \phi_i(v)w_j.$$ If we plug in $v_i$ we see that $$\left( \sum a_i^j \phi^i \otimes w_j \right)\left( v_i \right) = \sum_{j} a^j_i w_j.$$ That is, the map defined by $\sum a_i^j \phi^i \otimes w_j$ is just the map represented in the bases $v$ and $w$ by the matrix $(a^j_i)$.

In the other direction, given a map $T : V \rightarrow W$, represent it in the bases $v$ and $w$ by $a^j_i$ and match it to the element $\sum a_i^j \phi^i \otimes w_j$. Just like you expand a vector $w \in W$ in a basis $w_i$ as $w = \sum_j a^j w_j$, where the coefficients are scalars, you expand a map $T$ as $T = \sum a_i^j \phi^i \otimes w_j$ where the "coefficients" $\sum_{i} a_i^j \phi^i$ are functionals in $V^*$.

As usual, everything passes from the linear algebra setting to the vector bundle setting, and you identify a tensor $\sum a_i^j dx^i \otimes \frac{\partial}{\partial x^j}$ with a map between $TM$ and $TM$, given in the frame $(\frac{\partial}{\partial x^j})$ by the matrix $(a_i^j)$.

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Small correction: the bundle in the first line should be $T^*M\otimes TM$, not $T^*M\otimes T^*M$. – Jack Lee Nov 27 '12 at 15:46
I've corrected it. Thanks! – levap Nov 28 '12 at 10:31
Thanks for the answer, appreciate it. – soup Nov 28 '12 at 17:55