# Is $M \oplus e^1 = e^2$ , i.e. trivial?

I am trying to prove formally that the Mobius bundle,M, over $$S^1$$ when summed with the trivial rank $$1$$ bundle $$e_1$$ isn't the trivial bundle. In other words $$M\oplus e_1\neq e_2$$.

First we write the cocycles for M:

We cover $$S^1$$ with $$U_a,U_b$$, slightly enlarged semicircles. $$U_a \cap U_b=E_1 \sqcup E_2$$. Then for $$x \in E_1: g_{ba}=-1$$ and for $$x\in E_2:g_{ba}(x)=1$$.

This means that for $$M\oplus e_1$$ the cocycles $$\hat{g}_{ab}(x)$$ are $$\begin{bmatrix} -1 & 0\newline 0 & 1\ \end{bmatrix}$$ and $$\begin{bmatrix} 1 & 0\newline 0 & 1\ \end{bmatrix}$$

,which we call $$A$$and$$B$$,in $$E_1$$ and $$E_2$$ respectively ( so the cocycles are constant functions in each component). In order for $$M\oplus e_1=e_2$$ we need to find $$h_i:U_i\rightarrow GL_2(\mathbb{R})$$, $$i\in \{a,b\}$$such that for $$x\in E_1:$$ $$\text{Id}=(h_1)^{-1}Ah_2$$ and for $$x\in E_2:\text{Id}=(h_1)^{-1}Bh_2$$.

This forces one of the $$h_i$$ to change sigh of the determinant as $$x$$ goes for $$E_1$$ to $$E_2$$ which is impossible since we are considering real bundles. We are done.

First of all, I want to know if what i have proved is valid, regardless of my proof. If that is the case I would be grateful to comments on my proof, whether it is wrong or correct.

## 1 Answer

One more general way to prov this goes as follows. It is easy to see that the first Stiefel-Whitney class $$w_1(M) \neq 0$$, so $$w_1(M \oplus e_1)=w_1(M)+w_1(e_1)=w_1(M)$$ and $$w_1(e_2)=0$$ so these bundles are not isomorphic.

Your proof is also correct.