Let $T$ denote the $2$-torus, $P$ the projective plane, and $nT$/$nP$ the connected sum of $n$ tori/$n$ projective planes respectively.

1) how can I prove, that $nT$ and $nP$ are homeomorphic to their fundamental polygons, i.e. the quotients of the regular $4n$-gon / $2n$-gon, depicted below? I'm guessing induction on $n$. For $n=1$, the thing already holds, but how does one go through the inductive step? No need for actual formulas, just intuitively to the point when one can easily figure out the formulas himself. enter image description here

2) We can see that all points enter image description here become the same under the quotient projection. This gives $nT$ and $nP$ the CW-structure, depicted below, enter image description here $~~~~~~$ where we attach a disc $B^2$ via the attaching map $a_1b_1a_1^{-1}b_1^{-1}\ldots a_nb_na_n^{-1}b_n^{-1}=[a_1,b_1]\ldots[a_n,b_n]$ for $nT$, and $a_1^2\ldots a_n^2$ for $nP$.

By the theorem regarding the fundamental group of a CW-complex, it follows that $$\pi_1(nT)=\langle a_1,b_1,\ldots,a_n,b_n|[a_1,b_1]\ldots[a_n,b_n]\rangle~~~~\text{and}$$ $$\pi_1(nP)=\langle a_1,\ldots,a_n|a_1^2\ldots a_n^2\rangle.$$ Is this correct?

3) How can $nP$ and $nT$ be homeomorphic to their CW-decompositions, when at the point enter image description here, the locallly $2$-euclidean property does not hold (the way I see it)?

  • 2
    $\begingroup$ The result $\pi_1(nT)=\langle a_1,\dots, b_g\,|\,[a_1,b_1]\dots[a_g,b_g]\rangle$ is correct. Also, look more closely - the locally $2$-euclidean property does hold at the red point. $\endgroup$ – user8268 Jun 9 '11 at 7:20
  • 1
    $\begingroup$ For your first question, you need to use the definition of the connected sum. Basically, in your picture of $nT$, you cut along the line segment from the beginning of $a_1$ to the end of $b_1^{-1}$. You get a torus with a disc removed (this is the pentagon with a free side) and something which is a $(n-1)T$ with a disc removed by induction. When you glue them together along the cut you get exactly the connected sum you need. $\endgroup$ – Miha Habič Jun 9 '11 at 8:18
  • $\begingroup$ Marvellous, I see it now. Thank you. $\endgroup$ – Leon Jun 9 '11 at 8:39

Your Answer

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

Browse other questions tagged or ask your own question.