I am trying to prove that $\mathbb{R}P^n$ is orientable iff $n$ is odd. One way to do that is to calculate the homology of the space, and then use the (heavy?) theorem that states that a $n$-dimensioanl connected closed manifold $M$ has $n$-th homology

$$\tilde{H}_n(M) = \begin{cases} \mathbb{Z} & \text{M is orientable}\\ 0 & \text{M is not orientable} \end{cases} $$

Now, I do not want to use this theorem, but rather prove this directly (I would also say that the way I calculate the $n$-th group of $\mathbb{R}P^n$ actually depends on $\mathbb{R}P^n$ being orientable iff $n$ is odd). For that, here are some definitions:

  1. The way orientation at a point $p\in M$ is defined using homology groups is, showing that $H_n(M,M\setminus\{p\}) = \mathbb{Z}$, and a generator for this group is an imbeding of a $n$-dimensinal simplex around $p$ (corresponding to $1\in \mathbb{Z}$) or it's reflection (corresponding to $-1\in \mathbb{Z}$) - choosing between those two gives an orientation at the point. Now, a loop in $M$ is said to preserve the orientation if when going through it one time, the orientation at the base point is not changed.
  2. The orientation chracter is the map $\pi_1(M) \to \{1,-1\}$, sending a homotopy class of a loop to $1$ iff the loop is orientation preserving. $M$ is orientable iff it has trivial charcter, and this statement is what I want to use.

Now for my attempt:

We know that $S^n$ is an order-2 covering for $\mathbb{R}P^n$, so it's fundamental group is $\mathbb{Z}/2\mathbb{Z}$ (when $n\ge 2$, the case $n=1$ being trivial). An explicit generator for the group is the closed loop, coming from a curve on $S^n$ that connects two antipode points.

Now I know that the antipodal map on $S^n$ induces $(-1)^{n+1} \cdot \operatorname{Id}$ on the homology group (as it is a composition of $n+1$ reflection, each gives one change of sign, as proved in homoology theory using the naturality of the Mayer-Vietoris sequence).

Using this, I want to derive that this non-trivial loop on $\mathbb{R}P^n$ is mapped by the character to $(-1)^{n+1}$, but how can this be done?

  • 1
    $\begingroup$ Well, what is your definition of orientation character? $\endgroup$
    – Pax
    Jul 21, 2015 at 10:08
  • $\begingroup$ If you like to work in the smooth world, you could use math.stackexchange.com/questions/163502/… . $\endgroup$ Jul 21, 2015 at 10:11
  • $\begingroup$ I should also note that your Homology characterization is quite wrong. The only requirement you need is $H^{dim(M)}(M)=\mathbb{Z}$, not the vanishing of the lower ones. $H^1(S^1\times S^1)=\mathbb{Z}^2$ after all. $\endgroup$
    – Pax
    Jul 21, 2015 at 10:22
  • $\begingroup$ You are right, I meant to write something else. I edited it, and added the definition of the oreintation character $\endgroup$
    – Mike
    Jul 21, 2015 at 10:24
  • $\begingroup$ What do you mean when you say that a loop is "orientation preserving"? $\endgroup$ Jul 21, 2015 at 11:09

1 Answer 1


Here is a solution, actually not involving homotopy groups at all:

In the real projective space, we want to follow the orientation given in a base point, going through the nontrivial closed loop until coming back to the base point.

To do so, we take the loop in $S^n$ which is the half-circle between two antipodal points; since $S^n$ is orientable, what remains to check is the consistency of the orientation in a neighborhood of the two glued points. This we do using the antipodal map, which is precisely the gluing function. The antipodal map is a composition of $n+1$ reflection, and thus will preserve the orientation in the neighborhood of the glued point iff $n$ is odd - this is true since each reflection twists the orientation (as noted above).


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