I have seen and worked through the explicit computation of the Morse homology of the sphere and the torus (with signs and all). But trying it for $\mathbb{R}\mathbb{P}^2$ has lead me to dead ends. Is there a fully worked out computation for this projective plane (with signs and all, for $\mathbb{Z}$-coefficients)?

As a start, the Morse function to utilize is $f(x_1,x_2,x_3)=i(|x_1|^2+|x_2|^2+|x_3|^2)$, in homogeneous coordinates on $\mathbb{R}\mathbb{P}^2$. On each neighborhood $U_1,U_2,U_3$ ($U_i$ denotes the set of coordinates $(x_1,x_2,x_3)$ where $x_i\ne 0$) there is one critical point, namely $(1,0,0)$, $(0,1,0)$, $(0,0,1)$ respectively, of Morse-index 1,2,3 respectively. Maybe the homology computation will be easier with a different function.


In the case of $RP^2$ you define $f([x:y:z])=ax^2+by^2+cz^2$ for $a<b<c$. Here you think of $[x:y:z]$ as normalized so that $x^2+y^2+z^2=1$; note that $f$ is well-defined. In a coordinate charts $x\neq 0$ you see that the only critical point there is $[1:0:0]$. Similarly in other two charts you get the points $[0:1:0]$ and $[0:0:1]$. The rough picture of the flow enter image description here

You see that there are two flowlines from max to the saddle and and from saddle to min. We know that the max to saddle is of same sign and saddle to min is opposite sign because we know the answer; if we did not know it we could orient the descending manifolds, say counterclockwise for max, and ``counterclockwise" for saddle - to the right on the bottom and to the left on the top; and + for the min. We have no ambient orientation so we have to be careful. The orientation conventions (see for example Hutchings notes) say we should use the flow along the trajectory to transport the orientation of descending manifold and orient the $TM_{pq}$ so that $TD_p=TM_{pq}\oplus T\gamma \oplus TD_q$. For the maximum this gives + orientation for both flowlines; for saddle since $T\gamma$ has different oriantations, we get one + and one -.

  • 1
    $\begingroup$ If I am understanding correctly, then the maximum gets mapped under the Morse-differential to twice the saddle, implying that $H_2(\mathbb{R}\mathbb{P}^2;\mathbb{Z})\cong 0$. The saddle is sent to $(+1-1)$ times the minimum, and so generates $H_1(\mathbb{R}\mathbb{P}^2;\mathbb{Z})\cong\mathbb{Z}/2\mathbb{Z}$ where the 2-torsion comes from the image of the maximum. $\endgroup$ – Matt Fahrad Nov 12 '12 at 0:45
  • $\begingroup$ Mike - yes, that's right. $\endgroup$ – Max Nov 12 '12 at 7:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.