Homeomorphism linear subspace

[cianfa72]

TL;DR Summary

Homemorphism between a linear subspace of $\mathbb R^8$ and $\mathbb R^4$

Hi,

consider the Euclidean space $\mathbb R^8$ and the projection map $\pi$ over the first 4 coordinates, i.e. $\pi : \mathbb R^8 \rightarrow \mathbb R^4$.

I would show that the restriction of $\pi$ to the linear subspace $A$ (endowed with the subspace topology from $\mathbb R^8$) is homeomorphism with $\mathbb R^4$. $A$ is defined by

$$x_5 = x_1, x_6=x_2, x_7=x_3, x_8 = x_4$$
$\pi$ is continuous and open, its restriction to $A$ is continuous as well. What about openess in the subspace topology ?

My idea is to show that projection's restriction is open as well using open balls as topology basis in $\mathbb R^8$. The intersection of an open ball with $A$ can be either empty or a set such that its projection is open in $\mathbb R^4$. Therefore we have an open continuous bijection between $A$ and $\mathbb R^4$ hence they are homemorphic.

Does it make sense ? Thanks.

 

[pasmith]

The Euclidean topology on $\mathbb{R}^n$ is the topology induced by the Euclidean norm. The subspace topology on the (vector) subspace $A$ is that induced by the restriction of the norm to $A$, which in this case reduces to $$\|x\|_A = \sqrt{2} \|\pi|_A(x)\|_{\mathbb{R}^4}.$$ It follows that $$\|\pi|_A(x) - \pi|_A(y)\|_{\mathbb{R}^4} = \frac{1}{\sqrt{2}} \|x - y\|_{A}$$ from which in turn it follows that both $\pi|_A$ and its inverse are continuous with respect to those norms, and thus with respect to the topologies induced by them.

Indeed, any linear map between finite dimensional normed vector spaces is continuous with respect to those norms, and the inverse of a linear map (if it exists) is again a linear map.

 

[cianfa72]

Indeed, any linear map between finite dimensional normed vector spaces is continuous with respect to those norms, and the inverse of a linear map (if it exists) is again a linear map.

In this case the two finite dimensional normed vector spaces are the vector subspace $A$ with the restricted norm and $\mathbb R^4$ with its standard Euclidean norm (induced from Euclidean inner product). $\pi|_A$ is the linear map (with its inverse).

Now if you take the 3-sphere $\mathbb S^3$ in $\mathbb R^4$ the restriction of $(\pi|_A)^{-1}$ on it is homeomorphism with its image (both with their subspace topologies).

 

[jbergman]

TL;DR Summary: Homemorphism between a linear subspace of $\mathbb R^8$ and $\mathbb R^4$

Hi,

consider the Euclidean space $\mathbb R^8$ and the projection map $\pi$ over the first 4 coordinates, i.e. $\pi : \mathbb R^8 \rightarrow \mathbb R^4$.

I would show that the restriction of $\pi$ to the linear subspace $A$ (endowed with the subspace topology from $\mathbb R^8$) is homeomorphism with $\mathbb R^4$. $A$ is defined by

$$x_5 = x_1, x_6=x_2, x_7=x_3, x_8 = x_4$$
$\pi$ is continuous and open, its restriction to $A$ is continuous as well. What about openess in the subspace topology ?

My idea is to show that projection's restriction is open as well using open balls as topology basis in $\mathbb R^8$. The intersection of an open ball with $A$ can be either empty or a set such that its projection is open in $\mathbb R^4$. Therefore we have an open continuous bijection between $A$ and $\mathbb R^4$ hence they are homemorphic.

Does it make sense ? Thanks.

$A$ is a submanifold with a global chart map,
$$ \phi(x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8) = \pi(x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8)|_A = (x_1, x_2, x_3, x_4)$$

In other words, our chart map for $A$ is equal to a projection onto the first four coordinates of $\mathbb{R}^8$ restricted to $A$.

We define our topology on $A$ by defining the inverse image of open sets in $\mathbb{R}^4$ under $\phi$ to be open in $A$. This induces the subspace topology on $A$ because projection maps are open.

More formally, given an open set in $V \in A$, then $V = \phi^{-1}(W)$, with $W$ open. But $V = \pi^{-1}|_A(W) = U \cap A$ with $U$ open in $\mathbb{R}^8$ because $\pi$ is continuous on $\mathbb{R}^8$ and the definition of $\pi|_A$ as a restriction map. Therefore every open set in $A$ is of the form $U \cap A$.

To prove the converse we have to show that $U \cap A$ is always open. But $\pi(U) = W$ is open in $\mathbb{R}^4$ because projection maps are open. We know that $\pi^{-1}|_A(W) = V$ is open in $A$. $\pi^{-1}(W) \supseteq U$. Therefore $\pi^{-1}|_A(W) \supseteq U \cap A$. But $\pi|_A$ is 1-to-1 on $A$ and $\pi|_A(U\cap A) = W$. Therefore $V= \pi^{-1}|_A(W) = U \cap A$.

So we have constructed a topology on A that is both the subspace topology of $\mathbb{R}^8$ and with an explicit homeomorphism to $\mathbb{R}^4$.

 

[cianfa72]

To prove the converse we have to show that $U \cap A$ is always open. But $\pi(U) = W$ is open in $\mathbb{R}^4$ because projection maps are open. We know that $\pi^{-1}|_A(W) = V$ is open in $A$. $\pi^{-1}(W) \supseteq U$. Therefore $\pi^{-1}|_A(W) \supseteq U \cap A$. But $\pi|_A$ is 1-to-1 on $A$ and $\pi|_A(U\cap A) = W$. Therefore $V= \pi^{-1}|_A(W) = U \cap A$.

Sorry, why $\pi|_A(U\cap A) = W$ holds true ? If we applies the map $\pi|_A$ 1-to-1 on $A$ to both terms of $\pi^{-1}|_A(W) \supseteq U \cap A$ we get $W \supseteq \pi|_A (U \cap A)$.

 

[jbergman]

Sorry, why $\pi|_A(U\cap A) = W$ holds true ? If we applies the map $\pi|_A$ 1-to-1 on $A$ to both terms of $\pi^{-1}|_A(W) \supseteq U \cap A$ we get $W \supseteq \pi|_A (U \cap A)$.

Yeah, that is wrong. I need to fix that part of the proof.

 

[jbergman]

Yeah, that is wrong. I need to fix that part of the proof.

A simpler proof as follows. We give $A$ the subspace topology and again we have for our global chart map $\pi|_A$ which we know is continuous because restrictions of continuous maps are continuous. We also have the inverse map given by

$$\pi|_A^{-1}(x)=(x_1, x_2, x_3, x_4,x_1, x_2, x_3, x_4)$$

This is continuous because we can view $\mathbb{R}^8$ as $\mathbb{R}^4 \times \mathbb{R}^4$ with both $\pi_1 \circ \pi|_A^{-1}$ and $\pi_2 \circ \pi|_A^{-1}$ continuous maps and the $\pi_i$ projections onto the first and second components of $\mathbb{R}^4 \times \mathbb{R}^4$. Hence $\pi|_A^{-1}$ is continuous as viewed as a map into $\mathbb{R}^4 \times \mathbb{R}^4$ and $\mathbb{R}^8$. Hence it's restriction to $A$ is continuous.

 

[cianfa72]

Hence $\pi|_A^{-1}$ is continuous as viewed as a map into $\mathbb{R}^4 \times \mathbb{R}^4$ and $\mathbb{R}^8$. Hence it's restriction to $A$ is continuous.

You meant the restriction of the codomain/target of $\pi|_A^{-1}$ however it takes values only in $A$. Therefore for any $U$ open in $\mathbb R ^8$ $\pi_A(U \cup A) = \pi|_A(U)$ is open in $\mathbb R ^4$ since $\pi|_A^{-1}$ is continuous as map from $\mathbb R ^4$ to $\mathbb R ^8$.

 

[cianfa72]

You meant the restriction of the codomain/target of $\pi|_A^{-1}$ however it takes values only in $A$. Therefore for any $U$ open in $\mathbb R ^8$ $\pi_A(U \cup A) = \pi|_A(U)$ is open in $\mathbb R ^4$ since $\pi|_A^{-1}$ is continuous as map from $\mathbb R ^4$ to $\mathbb R ^8$.

Searching for it, the term for a such codomain restriction is corestriction. Basically we are corestricting the codomain of $\pi|_A^{-1}: \mathbb R^4 \rightarrow \mathbb R^8$ to its image $A$ then using on $A$ the subspace topology from $\mathbb R^8$.