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I am trying to show that $X= \Sigma^{m}_{i=1} \lambda_i\frac{\partial}{\partial x_i}$, where $\lambda:U \rightarrow \mathbb{R}$ is a smooth function is a vector field on $M$. However, as far as I can tell, if you plug $x \in M$ in, you get an element of $T_xM$ not $TM$. Where am I going wrong?

EDIT: As an aside from the original question, how would on go about proving that every vector field has this representation? I presume you use the fact that {$d_i$} is a basis, but I can't seem to formalise it.

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  • $\begingroup$ $TM = \displaystyle\bigsqcup_{x\in M} T_xM$ so everything works out $\endgroup$ – Max Apr 26 '18 at 18:09
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By definition, a vector field on $TM$ is a set containing one vector $v_x\in T_xM$ for each $x\in M$, so this is not a problem.

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