There's a very subtle step which I left to the exercises, and it was only when I wrote my own solutions manual and tried to do the exercise that I realize how tricky it was. I ended up rewriting the problem completely. @EricS
i think the couple of kids, who took the surfaces course from schlag last spring, in the current year who im TAing for saw cherns proof of gauss bonnet
Eric, I know I sent it to Balarka, but did I ever send you my paper on integro-geometric local Gauss-Bonnet/local Poincaré-Hopf? Neat stuff on polar varieties, too.
im excited to take a look at the complex geo exercises you sent me come fall though Ted, since i will probably be taking complex geo w Webster in all likelihood
@EricS: I did send you that paper, along with others that are more algebro-geometric. Plus all sorts of exercises on diff geo and complex geo and bundles & char. classes.
Let $\cdot : \mathbb{Z}\times \mathbb{Z}\rightarrow \mathbb{Z}$ be the usual multiplication and let $n\in \mathbb{N}$.
To show that $[a]\cdot_n [b]:=[a\cdot b]$ defines a concatenation of $\mathbb{Z}/n\mathbb{Z}$, we have to show that $[a]\cdot_n [b]\in \mathbb{Z}/n\mathbb{Z}$ for all $a,b\in \mathbb{Z}/n\mathbb{Z}$, or not?
Ah ok! So we have that $[a]=[a']$ means that $a\equiv a' \pmod n\Rightarrow \exists k_1\in \mathbb{Z}: a=a'+k_1n$. Respectively we have that $\exists k_2\in \mathbb{Z}: b=b'+k_2n$. Then $ab=(a'+k_1n)(b'+k_2n)=a'b'+(a'k_2+k_1b')n+k_1k_2n^2$. From that we get that $ab\equiv a'b'\pmod n$. This means that $[ab]=[a'b']$. Is everything correct?
If you take a connected component of a graph, that's a graph itself, but by handshake theorem the sum of the degrees is twice the the number of edges, in particular it's even
So you can't have a graph where only one vertex has odd degree
