Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $k,n$ be integers, $u_1,\dots,u_n \in U(k)$, $d_1,\dots,d_n \in \mathbb Z$ with $\sum_{i=1}^{n} d_i =: d \neq 0$.

Consider the map $w:= U(k) \to U(k)$ with $$w(v):= u_1 v^{d_1} \cdots u_n v^{d_n} \in U(k).$$

Since $U(k)$ is connected, each $u_i$ can be connected to $1_k \in U(k)$ and it is easily seen that $w$ is homotopic to $w'(u)= u^{d}$. It is also easy to see that the generic number of preimages of $w'$ is $|d|^k$ and this is also (up to sign maybe) equal to the degree of $w'$ as a map of oriented manifolds. In particular, as $d \neq 0$, the degree of $w'$ (and hence $w$) is non-zero. We conclude that $w$ is surjective and hence, there exists $v \in U(k)$ such that

$$u_1 v^{d_1} \cdots u_n v^{d_n} = 1_k \in U(k).$$

The argument above is due to Murray Gerstenhaber and Oskar Rothaus. I am asking for alternative proofs of this statement which do not rely on degree theory or algebraic topology.

More specifically, already the following is not so clear to me:

Question: Let $t \mapsto u_{i,t}$ be smooth paths which connect $u_i$ with $1_k$. Then, for each $t \in [0,1]$, there exists a solution to the equation $$u_{1,t} v^{d_1} \cdots u_{n,t} v^{d_n} = 1_k.$$ Is there a continuous (smooth) path of solutions $v_t$?

It seems that the answer must be yes.

Question: If the answer to the first question is yes, then it fairly easy to see that $v_t$ satisfies some first-order differential equation. Is there a way of constructing $v_t$ as the solution to this differential equation?

share|improve this question
add comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.