2
$\begingroup$

I'm having trouble understanding an example in Guillemin and Pollack. If $f:R^k\rightarrow R$ be defined by

$f(x) = |x|^2 = x_1^2+...+x_k^2$

The derivative $df_x$ at the point $a = (a_1,..,a_k)$ has matrix $(2a_1,..,2a_k)$. Thus

$df_a:R^k\rightarrow R$ is surjective unless $f(a)=0$, so every nonzero real number is a regular value of $f$. I can't understand how to check $df_a$ is surjective and why it is obvious that every nonzero real number is a regular value of $f$? A hint is appreciated. Thanks.

$\endgroup$
1
$\begingroup$

It is clear that if $f(a)=a_1^2+\cdots+a_k^2 = 0$, then $a=0$ which implies $df_a = (0,\dots,0)$ is the zero map (obviously not surjective).

If $f(a)\neq 0$, then at least there is one component of $a$ which is nonzero. Wlog, assume $a_1 \neq 0$. Therefore the components of $df_a$ not all zero. For any nonzero $x \in \Bbb{R}$, $$ df_a([x/2a_1, \, 0, \cdots ,\, 0]) = 2a_1 \frac{x}{2a_1} + 2a_2 \cdot 0 + \cdots + 2a_k \cdot 0 = x. $$ So $df_a$ is onto.

Since any nonzero real number is the image of a particular vector under all linear map $df_a$ (where $a$ is point such that $f(a) \neq 0$), therefore they're all regular value.

$\endgroup$
0
$\begingroup$

Hint: What is the rank of the linear map $2(a_1,\dots, a_k):\Bbb R^k\rightarrow \Bbb R$? (And, what is the dimension of $\Bbb R$?)

$\endgroup$
  • $\begingroup$ $dim(R)=1$. I'm not sure about the rank of the linear map. It is a row matrix and can have $k$ linearly independent elements? $\endgroup$ – manifolded Jan 31 at 8:56
  • $\begingroup$ The rank always equals the dimension of the column space, which equals that of the row space. So...? $\endgroup$ – Chris Custer Jan 31 at 9:01
0
$\begingroup$

For $a \neq 0$, the reason why $df_a$ is surjective because $df_a$ is a linear map from $T_p \mathbb{R}^k$ to $T_{f(p)} \mathbb{R}$ and since $\operatorname{rank} df_a = 1 = \operatorname{dim} T_{f(p)}\mathbb{R}$ it follows that $df_a$ is surjective.

To see that $\operatorname{rank} df_a = 1$, note that the matrix $[ 2a_1 \dots 2a_k]$ will always have some entry to be non-zero (since we assumed that $a \neq 0$ so some $a_i$ must be non-zero for some $0 \leq i \leq k$) and having one non-zero entry is all we need to conclude that the rank of this matrix is $1$.

Recall that the for a smooth map $F : M \to N$ between smooth manifolds, a point $c \in N$ is a regular value of $F$ if and only if either $c$ is not in the image of $F$ or at every point $p \in F^{-1}(c)$ we have $dF_p : T_pM \to T_{F(p)}M$ to be surjective.

So choose some non-zero real number $y$, if $y \not\in f[\mathbb{R^k}] $ then we're done otherwise suppose that $y \in f[\mathbb{R^k}]$, then $f^{-1}(y)$ is nonempty. Choose any $a \in f^{-1}(y)$, and then note that $a$ must clearly be non-zero in $\mathbb{R}^k$, thus by the above we have $df_a$ to be surjective and since $a$ was chosen arbitrarily here, this holds for all $a \in f^{-1}(y)$, and so it follows that every non-zero real number is a regular value for $f$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.