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I am trying to understand a line of reasoning in Guillemin and Pollack. Let $U\subset \mathbb{R}^k$ and $V\subset \mathbb{R}^l$ be open sets, and let $f:V\rightarrow U$ be smooth. Use $x_1,...,x_k$ for the standard coordinate functions on $\mathbb{R}^k$ and $y_1,...,y_l$ on $\mathbb{R}^l$.

I am trying to understand the following equation: $f^*dx_i=\sum_{j=1}^l\frac{\partial f_i}{\partial y_j}dy_j=df_i$, where $f^*$ denotes the pullback by $f$.

For a vector $v\in V$, we have $f^*dx_i(v)=dx_i(df(v))$, but I'm not sure where to go from here. I know that $df$ pushes the vector forward into $\mathbb{R^k}$, and that $dx_i$ in turn measures the $i$th coordinate of that vector; however, I am confused why it should be equal to the some on the right. How are they equal?

ponchan
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  • Are you sure that $f^dx_i(v)=dx_i(df(v))$? I would say the right equation is $f^dx_i(v)=df^*(x_i)(v)$ which is equal to $d(x_i\circ f)(v)=df_i(v)=\sum_{j=1}^l\frac{\partial f_i}{\partial y_j}dy_j(v)$ – Javi Apr 18 '18 at 21:53
  • The pullback in general is defined as: if $f:X\rightarrow Y$ is a smooth map, $f(x)=y$, and $w$ is a p-form on $Y$, then $f^w=(df_x)^w[f(x)]$ – ponchan Apr 18 '18 at 21:56
  • This might help you, is the same question but maybe with a different point of view https://math.stackexchange.com/questions/456586/show-fdx-i-sum-j-1l-frac-partial-f-i-partial-y-j-dy-j-df-i?rq=1 – Javi Apr 18 '18 at 21:59
  • I am confused why you can write $Y$ as a sum in that way. – ponchan Apr 18 '18 at 21:59
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    Because the $\frac{\partial}{\partial y_i}$ form a basis of the tangent space and $Y$ is a tangent vector. – Javi Apr 18 '18 at 22:01
  • What does it mean to raise the vector $Y$ to a power? – ponchan Apr 18 '18 at 22:03
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    It's not a power, it's just a common notation in this field to write the components, you can write them as subindices if you dare – Javi Apr 18 '18 at 22:05
  • Also, don't the $\frac{\partial}{\partial y_i}$ form a basis for the dual space, and not the tangent space itself? – ponchan Apr 18 '18 at 22:06
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    Strictly speaking you're right, it is implicit that $Y$ and $\frac{\partial}{\partial y_i}$ are being evaluated at some point. – Javi Apr 18 '18 at 22:09

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