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?
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$\begingroup$ How do you get from $2r_1+n\geq 2n$ to $r_1\geq n$? $\endgroup$– MaxJul 13, 2021 at 5:46
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$\begingroup$ On the other hand, in the case $x=y$ you have $2r_1+n-r_1\geq 2n$, so $r_1=n$, indeed. $\endgroup$– MaxJul 13, 2021 at 5:54
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1$\begingroup$ 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 $N$). $\endgroup$– MaxJul 13, 2021 at 5:58
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$\begingroup$ You're right, I realized I made a calculation error and edited the question without re-reading carefully. $\endgroup$– Michael L.Jul 13, 2021 at 6:11
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1$\begingroup$ @Vajra $N$ might not be equipped with a subtraction operation to allow you to make sense of $f(x)-f(y)$ $\endgroup$– AmrJul 13, 2021 at 12:19
1 Answer
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.
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$\begingroup$ 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. $\endgroup$ Jul 15, 2021 at 17:44
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$\begingroup$ @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. $\endgroup$– AmrJul 16, 2021 at 15:34