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In problem 7, pg. 75 of Guillemin and Pollack's Differential topology, there is a hint saying

The set $S\subset(\mathbb{R}^N)^\ell$ of linearly independent $\ell$-tuples in $\mathbb{R}^N$ is open in $\mathbb{R}^{N\ell}$ and the map $\mathbb{R}^\ell\times S\to\mathbb{R}^N$ defined by $$ F\colon [(t_1,\dots,t_\ell),v_1,\dots,v_\ell]\mapsto\sum t_iv_i $$ is a submersion.

Various errata online point out that this map is only a submersion away from $0$.

However, if the coordinate functions for $F$ are $F_j\colon\mathbb{R}^\ell\times S\to \mathbb{R}$ defined by mapping $[t_1,\dots,t_\ell,v_1,\dots,v_\ell]\mapsto \sum t_iv_i^{(j)}$, where I denote $v_i^{(j)}$ to be the $j$th entry of the $v_i$. Then isn't $\partial F_j/\partial t_i$ just $v_i^{(j)}$? Then in the matrix representation for $DF$, won't the leftmost $N\times N$ minor be that with rows equal to the $v_i$? But for any $\ell$-tuple of $v_i$, they are linearly independent by definition of $S$, so this minor is invertible, and $DF$ is surjective, so $F$ is a submersion.

Why would this fail at $0$? Am I computing the derivative wrong?

Clara
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1 Answers1

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According to the rule for differentiating a bilinear mapping the required differential is: $$DF_{[(t^0_1,\dots,t^0_\ell);(v_1^0,\dots,v_\ell^0)] } [(\delta t_1 ,\dots,\delta t_\ell);(\delta v_1,\dots,\delta v_\ell)] =\sum_ {i=1} ^{l} t^0_i\delta v_i+ \sum_ {i=1} ^l\delta t_iv_i^0 $$ This immediately shows that $F$ is a submersion unless we have both $(t^0_1,\dots,t^0_\ell)=(0\dots,0)$ and $l\lt N$.

Edit
As an answer to Clara's request in her comment let me remind whomever is concerned that the differential of a bilinear map $f:A\times B\to C$ between vector spaces at the point $(a,b)\in A\times B$ is the linear map $$Df_{(a,b)}:A\times B\to C: (u,v)\mapsto f(a,v)+f(u,b)$$ which I like to write for psychological-mnemonic reasons $$Df_{(a,b)}(\delta a,\delta b)= f(a,\delta b)+f(\delta a,b) $$

Georges Elencwajg
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  • I'm sorry, what are the $\delta$ preceding the $t_i$ and $v_i$? Also, what's the rule for differentiating a bilinear mapping? I thought I knew it but must be applying it wrong. Thanks. – Clara Feb 11 '15 at 17:58
  • $\delta t_i$ is an arbitrary real number. The notation is chosen for psychological reasons: it corresponds to the intuition of a small variation. I have displayed the rule for differentiating a bilinear mapping in an Edit. – Georges Elencwajg Feb 11 '15 at 18:59