I start to think of this is question when I attempt exercise 1.3.9 on Guillemin and Pollack's Differential Topology

Consider the projection $\varphi: \mathbb{R}^N \rightarrow \mathbb{R}^k$: $$(x_1, \dots, x_N) \mapsto (x_{i_1}, \dots, x_{i_k})$$ Differentiating: $$d \varphi: T_x(X) \mapsto \text{ span}(e_{i_1}, \dots, e_{i_k})$$

So just to confirm - differentiating $x_{i_1}, \dots, x_{i_k}$ result span($e_{i_1}, \dots, e_{i_k}$)?


A projection $\pi: \mathbb{R}^n \rightarrow \mathbb{R}^k$ defined by $$ \pi (x) = \sum_{j=1}^k [x \cdot e_{i_j}]e_j $$ will push $\frac{\partial}{\partial x^{i_j}}$ to $\frac{\partial}{\partial y^{i_j}}$ under the differential $d\pi$. Let $y$ be the coordinate system $(y^{i_j})$ for $j=1, \dots k$. Consider: $$ d\pi (\frac{\partial}{\partial x^{i_j}}) = \sum_{l=1}^k \frac{\partial y^{i_l}}{\partial x^{i_j}}\frac{\partial}{\partial y^{i_l}}$$ where $y = \pi (x)$ hence $y^l = x \cdot e_{i_l}= x^{i_l}$ and $\frac{\partial y^l}{\partial x^{i_j}}=\frac{\partial x^{i_l}}{\partial x^{i_j}} = \delta_{i_j,i_l} = \delta_{jl}$ thus, $$ d\pi (\frac{\partial}{\partial x^{i_j}}) = \sum_{l=1}^k\delta_{jl}\frac{\partial}{\partial y^{i_l}} = \frac{\partial}{\partial y^{i_j}}$$ I think you want to identify $\frac{\partial}{\partial y^{i_j}}$ with $e_j$. Under that assumption I suppose your claim is true.

On the other hand, if you wish to view that span as the copy of $\mathbb{R}^k$ embedded in $\mathbb{R}^n$ in the natural manner by setting the complement of the $x^{i_j}$ coordinates to zero then the span is literally accurate. However, I'm not sure what you intend so I wrote this post. I suppose, the formula for the embedded case is just $$ \pi (x) = (0,...0,x^{i_1},0,...,0,x^{i_k},0,...0) \in \mathbb{R}^n. $$

  • $\begingroup$ Fantastic, thanks so much James! Let me read it carefully!!! $\endgroup$ – WishingFish Jul 3 '13 at 2:06
  • $\begingroup$ @WishingFish let me know if something needs further detail. For example, I suppose technically I'm suppressing the point-dependence of the differential because it's everywhere and not too interesting here... $\endgroup$ – James S. Cook Jul 3 '13 at 2:37
  • $\begingroup$ ok - still reading it... -_- $\endgroup$ – WishingFish Jul 3 '13 at 2:38
  • $\begingroup$ Your notation is very physical - I like it but have some trouble understand it. At the beginning, what do you mean why $e_{ij}$? Thank you very very much. $\endgroup$ – WishingFish Jul 3 '13 at 2:40
  • $\begingroup$ @WishingFish let me explain, $x \cdot e_{i_j}$ (the dot-product selects the $i_j$-th cartesian component of $x$ then multiplying by the standard basis element $e_j \in \mathbb{R}^k$ places it in the $j$-th component of $\mathbb{R}^k$. This for me is the main issue, if you say the projection goes to $\mathbb{R}^k$ then this forces a non-standard coordinate system on $\mathbb{R}^k$. Not a big deal, it's just a relabeling of standard numbered $1,2,...,k$ with $i_1,i_2,...i_k$. $\endgroup$ – James S. Cook Jul 3 '13 at 11:19

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