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Consider the projective space $\mathbb R P^n$, constructed as the quotient space of the sphere $S^n \subseteq \mathbb R^{n+1}$ with the equivalence relation where we identify opposite points.

I now want to prove or disprove that the set $M := \{(x, [y]) : [y] \in \mathbb R P^n, x \in [y]\}$ is a submanifold of the manifold $N := \mathbb R^{n+1} \times \mathbb R P^n$.

Now I rather suspect that it's not a submanifold, but I wasn't able to prove it so far. My thinking was, $M$ is a submanifold of $N$ if and only if there is an injective immersion $\iota: M \to N$. Now the first choice for such an immersion would be the natural map $M \to N, (x, [y]) \mapsto (x, [y])$. To check if that's an immersion, we would need to check that the Jacobean matrix is injective. I have not proven it yet but I suspect it will not be injective, basically because the components of a vector $x$ of the first component of $(x, [y])$ also appear in the $[y]$-component so the Jacobean it shouldn't add up to full rank?

But let's assume that's true, then I would have shown that the identity map $M \to N$ isn't an immersion... how would I go about showing that there doesn't exist any other map $M \to N$ that happens to be an injective immersion? Is there an easy way to see on why this is or isn't the case?

The most similar question I could find was this one and the set in question is not the same there as far as I can tell.

moran
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  • As stated, your question is ill-defined. Recall that given a manifold $M$, a subset $N$ together with a topology and a smooth structure is called an immersed submanifold of $M$ if the inclusion map is a smooth immersion. You haven't specified which topology and which smooth structure you put on $N$ and without this, you can't even check the inclusion map (or any other map) is a smooth immersion. – levap Nov 05 '17 at 09:04
  • What I suspect you want to do is to show that $N$ is an embedded submanifold which will mean that with respect to the subspace topology, there will be a (unique) smooth structure such that the inclusion is a smooth embedding. In fact, $N$ will then be diffeomorphic to $S^n$. – levap Nov 05 '17 at 09:10
  • @levap Thank you for pointing that out; yes, I think that's what I want to show, sorry for the confusion from the question. What would that unique smooth structure be? Could I just take the $S^n \times {c} \subseteq \mathbb R^{n+1} \times P \mathbb R^n$ and the inclusion map $\pi: (x, [x]) \mapsto (x, c)$ which would then be trivially smooth (I think)? – moran Nov 05 '17 at 11:31

1 Answers1

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Yes it is a submanifold, since the projection $M \to \Bbb RP^n$ is a line bundle. You can trivialize this vector bundle easily on the usual open set $U_i = \{[y_0: \dots : y_n] \in \Bbb RP^n : y_i \neq 0\}$. This will give you a local parametrization of $M$, and this shows that $M$ is a submanifold of $N$.

Nicolas Hemelsoet
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