# Tangent space to $\mathbb{R}P^{n}$

I could not find any other question here related to this. If I have missed out, then this could be voted as a duplicate(Sorry if it is!).

I was just trying to figure out the tangent space to the projective space and intuitively I thought that as the space is itself obtained by identifying antipodal points on the sphere, one could look at the tangent space to this quotient space as quotient of the tangent spaces to the sphere at the antipodal points.

In other words, if $p \in \mathbb{R}P^n$ is obtained by identifying $x,-x \in \mathbb{S}^n$, then does $T_p(\mathbb{R}P^n)$ stem out of identifying the spaces $T_x(\mathbb{S}^n)$ and $T_{-x}(\mathbb{S}^n)$??

Is this "quotient of tangent spaces" being the same as the tangent space of the quotient space true in general??

On a related note, I came across a problem instructing one to show that if

$$\pi : \mathbb{S}^n \to \mathbb{R}P^n$$ is the quotient map with $\pi(x) = p$, then there is an isomorphism $$\pi_*|_x :T_x(\mathbb{S}^n) \to T_p(\mathbb{R}P^n)$$.

How does one show this?? I can show that $\pi$ is smooth, but not sure whether it is a diffeomorphism, whence the above result would follow ??

I hope I have not missed out on something obvious here. Thanks for your patience.

• What exactly is your definition of tangent space? Also, $\pi$ is smooth, but is not a diffeomorphism because it is not $1-1$. None the less, it is a local diffeomorphism: Around every point in $S^n$ there is a neighborhood $U$ for which $\pi|_U$ is a diffeo onto its image. This is enough to prove that $\pi_\ast$ is an isomorphism. (I believe it's also necessary, but I may be missing a hypothesis or two) – Jason DeVito Sep 22 '14 at 14:04
• Extremely sorry for late reply. Network connection is a bit erratic here in India. I have taken the tangent space at any point of the sphere as merely the set of all orthogonal vectors to the radial line passing through that point. Yes local diffeomorphism is actually what I wanted to show too. A glaring omission, still. – Vishesh Sep 22 '14 at 15:06

## 1 Answer

I would suggest you to read the Example starting on the end of p. 10 here http://www.math.cornell.edu/~hatcher/VBKT/VB.pdf

Also in Milnors "Characteristic Classes" he computes the tangent space of the projective space in Chapter 2 or 3 I think. I strongly suggest you this one!

To answer your other question: Since the antipodal action is smooth, you get a smooth covering space to the quotient $S^n \to RP^n$, hence a local diffeomorphism. In particular induced maps on tangent spaces are isomorphisms.

• Thanks a lot. I will take a look at Milnor's book. I have very little idea about bundles. I am still a starter in Differential Topology. But does this apply to all quotient manifolds in general?? – Vishesh Sep 22 '14 at 15:09
• Milnor's book will also give you a general good idea for manifolds and vector bundles! So go for it! Of course not to all kind of quotients, but if you have a properly discontinous smooth action on a manifold, then yes. – Daniel Valenzuela Sep 22 '14 at 15:12
• Ok. Lastly, is it possible to show the local diffeomorphism without resorting to covering space notions??Thanks a lot. – Vishesh Sep 22 '14 at 15:14
• Ok, I got the reasoning for the diffeomorphism and the Milnor book is really a good read, cheers!! – Vishesh Sep 22 '14 at 17:16