# Characterizing the smooth maps $f : M\to N$ such that $E = \{(x, y)\in M\times M : f(x) = f(y)\}$ is a smooth submanifold of $M\times M$.

Let $$M$$ and $$N$$ be smooth manifolds. Let's say that I want characterize the smooth maps $$f : M\to N$$ such that $$E = \{(x, y)\in M\times M : f(x) = f(y)\}$$ is a smooth submanifold of $$M\times M$$ by the implicit function theorem. The implicit function theorem states that if $$W\subseteq N$$ is a smooth submanifold with $$\dim(W) = k$$, $$f$$ is transverse to $$W$$, and $$f^{-1}(W)\neq \emptyset$$, then $$f^{-1}(W)\subseteq M$$ is a smooth submanifold of codimension $$k$$. Let \begin{align*} F : M\times M\to{}& N\times N \\ (x, y)\mapsto{}& (f(x), f(y)) \end{align*} and $$D$$ be the diagonal in $$N\times N$$. Then, the transversality condition of $$F$$ to $$D$$ is stated as $$\mathrm{d} F(x, y)(T_{x, y}(M\times M))+T_{f(x), f(y)}D = T_{f(x), f(y)}(N\times N)$$ for all $$x, y\in M$$ such that $$f(x) = f(y)$$. We note that $$T_{f(x), f(y)}(N\times N) = T_{f(x)}N\times T_{f(x)}N$$ and that $$T_{f(x), f(y)}D = \{(v, v) : v\in T_{f(x)}N\}$$ Then, we write $$\mathrm{d} F(x, y)(T_{x, y}(M\times M)) = \{(v, w) : v\in \mathrm{d} f(x)(T_xM), w\in \mathrm{d} f(x)(T_yM)\}$$ It seems to me that for $$\dim(N) > 1$$, the transversality condition must be met iff $$f$$ is a submersion, as if we take $$x = y$$ and let $$r$$ be the dimension of $$\mathrm{d} f(x)(T_xM)$$, we must have $$2r+n-r\geq 2n$$ or $$r\geq n$$ for the transversality condition to be satisfied at $$(x, x)$$, which implies $$r = n$$. Therefore, if $$f$$ is a submersion, then $$E = F^{-1}(D)$$ is a smooth submanifold of $$M\times M$$ of codimension $$n$$. Is this in fact correct? Also, is there a weaker condition we can put on $$f$$ for this statement to be true?

• How do you get from $2r_1+n\geq 2n$ to $r_1\geq n$?
– Max
Jul 13, 2021 at 5:46
• On the other hand, in the case $x=y$ you have $2r_1+n-r_1\geq 2n$, so $r_1=n$, indeed.
– Max
Jul 13, 2021 at 5:54
• If the image of $f$ is a submanifold of $N$ and $f$ submerses onto that, then $E$ will also be smooth (after all, it can't tell the difference between the image of $f$ and $N$).
– Max
Jul 13, 2021 at 5:58
• You're right, I realized I made a calculation error and edited the question without re-reading carefully. Jul 13, 2021 at 6:11
• @Vajra $N$ might not be equipped with a subtraction operation to allow you to make sense of $f(x)-f(y)$
– Amr
Jul 13, 2021 at 12:19

Answering your question whether we can put a weaker condition to guarantee that $$E$$ will be a submanifold of $$M$$: Yes, I believe the requirement that rank of derivative map of $$f$$ is constant will imply that $$E$$ is a submanifold of $$M$$, because by the constant rank theorem $$f$$ will look like a linear map in suitable coordinates around any point and so in suitable coordinates $$E$$ will look like a linear space and so it will be a submanifold. Note: I didn't write everything down with all details so there is a minor chance I could be wrong.
• Is the idea that the charts guaranteed by the constant rank theorem on $M\times M$ can be restricted to $E$? I don't actually quite see how to make this work, although it's an intuitive idea. If you'd be willing to expand on this, I'd appreciate it. Jul 15, 2021 at 17:44
• @MichaelL. Using the coordinate system provided by the constant rank theorem, the pair $(E,M)$ will be appear in these coordinate system as a pair $(\mathbb{R}^{2dim(M)},V)$ where $V$ is a linear space. Thus, $E$ is locally diffeomorphic to $V$ which is a manifold. Thus, $E$ itself is a manifold. So yes, the restriction to $E$ happens naturally.