MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For handle decomposition of surface, suppose I have a twisted 1-handle(twisted only once) adjacent to an isolated pair of linked handles, is the handle slide operation enough to move the previous one to become 3 twisted 1-handle(each is the same as the previous twisted 1-handle), which proves that these two are homeomorphic to each other?

share|cite|improve this question
Let $P$ be the real projective plane and $T$ the two-torus. Let $\#$ denote connect sum. You are asking how one proves that $P \# T$ is homeomorphic $P \# P \# P$, correct? ---- If that is the case you can find the answer in Jeff Weeks book "Shape of space". However it might be more useful to you to start drawing pictures of what handle slides do to the arrangement. – Sam Nead Apr 20 '11 at 8:55
Yes, the handle slide operation is enough to show this. – Jim Conant Apr 20 '11 at 18:35
Any reference I can look at? There are too few materials online about this topic. – Hu Yi Chen Apr 20 '11 at 19:24
up vote 3 down vote accepted

I'm too lazy to draw this right now, but I think I can describe it anyway. Consider a twisted band attached right next to a dual pair of untwisted bands. I'll write this $A\bar{A}BCBC$. Here the overbar represents a twist and the bands attach to the same letters on each end. Now slide the B band into the middle of the A band to get $AB\bar{A}C\bar{B}C$. Slide the left foot of the $C$ band over the B band into the middle of the A band to get $ABC\bar{A}\bar{C}\bar{B}$. Now slide the right end of the $A$ band along the connected boundary until it is adjacent to the left end of the $A$ band. It slides once over the $B$ and $C$ bands, so picks up two half-twists. So we get $A\bar{A}BC\bar{C}\bar{B}$. Now take the entire $C$ band and slide it along the connected boundary until it lies next to the $B$ band on the right, yielding the desired $A\bar{A} B\bar{B} C\bar{C}$. Ta da!

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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