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let $X$ be a $n$-manifold. let $A=\{(x,y,z) \, |\,x=y\}$. I want to see if $A$ is a submanifold of $X^3$.

Consider the map $\Delta\times 1:X\times X\rightarrow X \times X\times X;\, (x,y)\mapsto (x,x,y)$.

If $U_x$ and $U_y$ are neighborhoods of $x$ and $y$ in $X$ then $\Delta\times 1 (U_x\times U_y)=\Delta(U_x)\times U_y$ is a neighborhood of $\Delta(x,y)=(x,x,y)$ in $X^3$, and we can see that $\Delta(U_x)\times U_y$ is in $A$. So this is a neighborhood of $(x,x,y)$ in $A$.

Now $U_x\cong \mathbb R^n$ and $\Delta(U_x)\cong U_x\cong \mathbb R^n$ and so $A$ is an $2n$-manifold.

Is my argument correct?and does it matter that the diagonal $\Delta(U_x)$ is a closed set, I mean don't we always want a neighborhood to be open?

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up vote 1 down vote accepted

Your argument is incorrect, though the conclusion that $A$ is a $2n$-manifold is true.

You claim without proof that $\Delta\times 1 (U_x\times U_y)=\Delta(U_x)\times U_y$ is a neighbourhood of $\Delta\times1(x,y)=(x,x,y)$ in $X^3$; it isn't. (Note that the "$\mathop\times1$" is missing in the question.) This is not because a neighbourhood must be open; it must merely contain an open set containing the point, but this set doesn't.

It is, as you say, a neighbourhood of $(x,x,y)$ in $A$, but you'll have to show that by other means. Also you don't provide an argument why $\Delta(U_x)\cong\mathbb R^n$ (which is correct).

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1)i thaught that if $f:X\rightarrow Y$ is continuous and $U$ is a neighb of $x$ then $f(U)$ is a neighb of $f(x)$ that's what i did for $f=\Delta\times 1$. 2) i don't understand what you mean "$ \times 1$ is missing in the question" 3)for all $Z$, the map $\Delta : z\mapsto (z,z)$ is a homeo from $Z$ to $\Delta(Z)$ , in particular when $Z=U_x\cong \mathbb R^n$ – palio Jul 9 '11 at 20:16
The continuous map f: $\mathbb R \rightarrow \mathbb R$ given by $f(x)=x^2$ takes the open set (-1,1) to [0,1), which is not open. You need f to be a homeomorphism for f to send open sets to open sets. You may be able to weaken the condition if f is differentiable and f' is not 0 in U. – gary Jul 9 '11 at 20:46

Your work could be simplified if you rotated your space by $45^o$ about the z-axis, after which it would have coordinates {(x,y,0)}; rotation is a homeomorphism, after which you get the x-z plane embedded in $\mathbb R^3$. Then your submanifold charts are already made for you, i.e., you have a map/chart that takes points in S to points in (x,y,0), by composing with the rotation.

Edit: I'm sorry to have to agree with others in that you do not seem to have a clear idea on the definition of submanifold. The one I use (which seemed the easiest in this case) is that a subset S of a manifold M is a submanifold of M if it can be given charts {$(f_1,U_i)$} such that $f(U_i\cap S)=(x_1,x_2,.,0.,0)$. In this case, you can get these charts by rotating about the z-axis by $45^o$, and then just using the identity, or standard embedding of $R^k$ in $R^m$

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thanks alot gary! actually i prefer a neighborhood argument because i'm working on a more general problem and just gave this as an example to understand the more general setting. – palio Jul 9 '11 at 20:22
gary it is clear for me that $A$ is homeomorphic to $X^2$ so it is a manifold. but as i said i'm working on more general problem so i need to show this by a neighborhood argument. – palio Jul 9 '11 at 20:46
palio: I am not clear on what you mean by a neighborhood argument. – gary Jul 9 '11 at 20:49
i mean to show that every point in $A$ has a neighborhood in $A$ that is homeo to $\mathbb R^{2n}$ – palio Jul 9 '11 at 20:55

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