2
$\begingroup$

Let $F$ and $G$ be smooth maps between smooth manifolds $M$ and $N$. Denote their differentials at $x\in M$ by $DF_x$ and $DG_x$. Why is the set $\{x\in M:F(x)=G(x), DF_x=DG_x\}$ closed in $M$?

I know of the result that if $f$ and $g$ are continuous maps between topological spaces $X$ and $Y$ with $Y$ being Hausdorff then $\{x\in X:f(x)=g(x)\}$ is closed in $X$. Unfortunately it doesn't seem like I can directly apply this result since I have $DF_x$ and $DG_x$ which are now maps between tangent spaces.

  • 0
    @NateEldredge I don't see why $DF$ and $DG$ map into $T^*N$. If they did then $DF_x\in T^*N$ which would mean that $DF_x$ takes in a tangent vector on $N$ and produces a real number but that's not the case. Please correct me if I'm wrong. – 2017-01-25
  • 0
    Oops, sorry. Deleted. – 2017-01-25
  • 0
    The easiest thing is probably to show the complement is open. You can break this up into the two cases where $F(x) \ne G(x)$, then you are done by the result you quoted and if $DF_x \ne DG_x$ but $F(x)=G(x)$ you can look at a chart containing $x$ and a chart containg $F(x)$ and use the continuity of the derivative of each function. – 2017-01-25

2 Answers 2

0

Take a sequence $x_n\to x$ and note that if $$ F(x_n)=G(x_n)\quad\text{and}\quad DF_{x_n}=DG_{x_n} $$ for all $n$, then $$ F(x)=G(x)\quad\text{and}\quad DF_{x}=DG_{x}. $$

  • 0
    $DF_{x_n}\in L(T_{x_n}M, T_{F(x_n)}N)$ while $DF_{x_m}\in L(T_{x_m}M, T_{F(x_m)}N)$ so they lie in different spaces when $n\neq m$. I don't see what it means for $DF_{x_n}$ to 'converge' to $DF_x$. – 2017-01-25
  • 0
    Sure, but how do you define continuity on the Grassmannian? Since $F$ is $C^1$, the convergence follows from continuity (if you want, using the canonical distance on the Grassmannian). – 2017-01-25
-1

It suffices to prove that if $f: M \to N$ is smooth, then

$\{ x \in M : f(x) = 0 \text{ and } Df_x = 0 \}$

is closed in $M$ (just consider $f = F - G$ and $N$ embedded in some Euclidean space). Now,

$\{ x \in M : f(x) = 0 \text{ and } Df_x = 0 \} = \{ x \in M : f(x) = 0\} \cap \{ x \in M : Df_x = 0\}.$

Since $f$ is continuous, $f^{-1}(\{ 0\})$ is closed in $M$. Since $f$ is smooth, $Df : TM \to TN$ is also continuous as a map between tangent bundles. So, the map $E : M \to \text{Hom}(TM, TN)$ given by

$E(x) = Df_x, \quad x \in M,$

is also continuous. Thus, $E^{-1}(\{0\})$ is also closed in $M$. Hence,

$\{ x \in M : f(x) = 0 \text{ and } Df_x = 0 \} = f^{-1}(\{0\}) \cap E^{-1}(\{0\})$

is closed, as required.