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The diagonal $Q$ in $X\times X$ is the set of points of the form $(x,x)$. Show that $Q$ is diffeomorphic to $X$, so $Q$ is a manifold if $X$ is.

Can anyone please help me to solve this question I have been trying to solve it since two days but I could not.

Thanks.

Zev Chonoles
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user63823
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  • A good place to start would be: do you understand how the smooth structure on $X\times X$ is defined from the smooth structure on $X$? – Zev Chonoles Feb 24 '13 at 21:08
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    What would you think is the most natural map $X\rightarrow X\times X$ which has image equal to $Q$? – Dan Rust Feb 24 '13 at 21:11
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    The wording of the exercise is a bit odd: it only makes sense to say that $Q$ is diffeomorphic to $X$ once we know that $Q$ is a smooth manifold, no? – Rasmus Feb 24 '13 at 21:28
  • The map $i:X\to X\times X$ given by $x\mapsto (x,x)$ is a smooth embedding. Then $Q=i(X)$ is an embedded submanifold of $X\times X$ with the property that $i$ is a diffeomorphism from $X$ to $i(X)=Q$. (Needs proof.) – wj32 Feb 24 '13 at 21:50
  • Can you well specify the data of your problem? What is $ X$? and what do we have to prove? –  Feb 27 '13 at 15:57

1 Answers1

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Since the term diffeomorphism applies by definition only to smooth manifods, the statement - as stated - makes little sense. Instead one can show the following

Proposition 1. Let $X$ be a topological space and $Q\subseteq X\times X$ the diagonal (endowed with the subspace topology of the product topology). Then the canonical embedding $i\colon x\mapsto (x,x)$ is a homeomorphism $X\to Q$.

Proof. The map $i$ is the unique continuous map $X\to X\times X$ that the universal property of the direct product guarantees for the identity $X\to X$ on both factors. On the other hand, the restriction to $Q$ of the canonical projection of $X\times X\to X$ (for either factor) is obviously inverse to $i$. $_\square$

In other words, $Q$ has exactly the same topological properties as $X$ has, especially

Corollary. $Q$ is a manifold if and only $X$ is. $_\square$

Proposition 2. Assume $X$ is a smooth manifold. Then the homeomorphism $i$ from proposition 1 is a smooth embedding (and hence $i$ is a diffeomorphism $X\to Q$).

Proof. Let $n=\dim X$, so $\dim(X\times X)=2n$. It suffices to show that for each $(x,x)\in Q$ there exists a chart $\phi\colon U\to \mathbb R^{2n}$ in the smooth atlas of $X\times X$ (as obtained for the product space from the given smooth atlas of $X$) such that $(x,x)\in U\subseteq X\times X$ and $\phi(U\cap Q)$ is the intersection of an $n$-dimensional plane with $\phi(U)$. To do so, pick a chart $\psi\colon V\to \mathbb R^n$ from the smooth atlas of $X$ for which $x\in V$. Then $U:=V\times V$ is open in $X\times X$ and by the definition of smooth structure on $X\times X$, the map $\phi:=\psi\times \psi\colon U\to \mathbb R^{2n}=\mathbb R^n\times \mathbb R^n$, $(\xi,\eta)\mapsto (\psi(\xi),\psi(\eta))$ belongs to the smooth atlas of $X\times X$. As $\phi(U\cap Q)$ is the diagonal plane $\mathbb R^n\times \mathbb R^n$, the claim follows. $_\square$

Hagen von Eitzen
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