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The suspension $SX$ of a topological space $X$ is defined as follows: $${\displaystyle S(X)=(X\times I)/\{(x_{1},0)\sim (x_{2},0){\mbox{ and }}(x_{1},1)\sim (x_{2},1){\mbox{ for all }}x_{1},x_{2}\in X\}}.$$ As an easy observation, the suspension of $\mathbb{R}P^{1}$, the projective space of dimension 1, is homeomorphic with $\mathbb{S}^2$. What happen for other dimension? Is there any well-known space that the suspension of $\mathbb{R}P^{n}$ is homeomorphic with that, for $n\geq 2$. Or, what we can say about homotopy groups of $S(\mathbb{R}P^{n})$? Especially, I am interested about calculation of $\pi_{3}(S(\mathbb{R}P^{3}))$.

Comment: We can calculate the homology groups of $S(\mathbb{R}P^{n})$, as for any topological space we have $\widetilde{H}_{i+1}(S(X))= \widetilde{H}_{i}(X)$.

MathFun
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As Mariano points out there is not much that can be said in general. It is these spaces which are the well-known spaces, and frequently appear as wedge summands in stable decompositions of other spaces. In low dimensions, however, we have the following special cases.

As you point out, $\mathbb{R}P^1\cong S^1$, so this is known.

$\mathbb{R}P^2\simeq M(\mathbb{Z}_2,1)$ is a mod 2 Moore space, as are all of its suspensions $\Sigma^n\mathbb{R}P^2\simeq M(\mathbb{Z}_2,n+1)$. See Jie Wu's monograph "Homotopy theory of the suspensions of the projective plane" for specific information on the homotopy of these spaces.

Finally we have $\mathbb{R}P^3\cong SO(3)$ is a a compact Lie group, so its top cell is stably trivial. In fact Naylor in his paper "Multiplications on $SO(3)$" shows that this happens after two suspensions, $\Sigma^2\mathbb{R}P^3\simeq S^5\vee \Sigma^2\mathbb{R}P^2$, and it is quite easy to see that this is the minimum number of suspensions for this to happen.

The space $\mathbb{R}P^7$ is also and H-space, so the top cell will split off at some point so give a stable equivalence $\mathbb{R}P^7\simeq_SS^7\vee \mathbb{R}P^6$. Unless someone tells me otherwise I do not believe the exact number of suspensions (or even good bounds) on the minimum number of suspensions for this splitting to occur is currently known.

Tyrone
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  • Thanks for your time. Is there any result for $\pi_3$ of $S(\mathbb{R}P^3)$? – MathFun Dec 28 '17 at 19:34
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    There is a cofibration sequence $S^3\rightarrow \Sigma \mathbb{R}P^2\rightarrow \Sigma\mathbb{R}P^2$ in which the spaces are sufficiently conneceted so as to give a Blakers-Massey exact homotopy sequence starting $\pi_3S^3\rightarrow\pi_3\Sigma\mathbb{R}P^2\rightarrow \Sigma\mathbb{R}P^3\rightarrow 0$. From Wu's monograph we have $\pi_3\mathbb{R}P^2=\mathbb{Z}_4$, and since $\mathbb{R}P^3$ does not split after one suspension the first map in the sequence is non-trivial. Therefore either $\pi_3\Sigma\mathbb{R}P^3=0$ or $\pi_3\Sigma\mathbb{R}P^3=\mathbb{Z_2}$. – Tyrone Dec 28 '17 at 21:35
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    Now the Freudental suspension theorem states that the suspension homomorphism $\pi_3\Sigma\mathbb{R}P^3\rightarrow\pi_4\Sigma^2\mathbb{R}P^3$ is surjective. However $\pi_4\Sigma^2\mathbb{R}P^3\cong\pi_4(\Sigma^2\mathbb{R}P^2\vee S^5)\cong\pi_4\Sigma^2\mathbb{R}P^2=\mathbb{Z}_2$, using Naylor's and Wu's results. Therefore we must have $\pi_3\Sigma\mathbb{R}P^3\cong\mathbb{Z}_2$. – Tyrone Dec 28 '17 at 21:38
  • Thank you very much for your complete explanation! I really appreciate it. – MathFun Dec 29 '17 at 06:04