TITLE: Using linear algebra to classify vector bundles over P^1
QUESTION [43 upvotes]: There is a theorem of Grothendieck stating that a vector bundle of rank $r$ over the projective line $\mathbb{P}^1$ can be decomposed into $r$ line bundles uniquely up to isomorphism. If we let $\mathcal{E}$ be a vector bundle of rank $r$, with $\mathcal{O}_X$ the usual sheaf of functions on $X = \mathbb{P}^1$, then we can write our line bundles as the invertible sheaves $\mathcal{O}_{X}(n)$ with $n \in \mathbb{Z}$. Thus, the decomposition can be stated as $$\mathcal{E} \cong \oplus_{i=1}^n \mathcal{O}(n_i) \quad n_1 \leq ... \leq n_r.$$
If we use the usual open cover of $\mathbb{P}^1$ with two affine lines $U_0 = \mathbb{P}^1 - \{\infty\}$ and $U_1 = \mathbb{P}^1 - \{0\}$, note that $\mathcal{O}_{U_0 \cap U_1} = k[x,x^{-1}]$ (with $\mathcal{O}_{U_0} = k[x]$ and $\mathcal{O}_{U_1} = k[x^{-1}]$). A vector bundle (up to isomorphism) $\mathcal{E}$ of rank $n$ is then a linear automorphism on $\mathcal{O}_{U_0 \cap U_1}^r$ modulo automorphisms of each $\mathcal{O}_{U_i}^r$ for $i = 0,1$. (I am looking at the definition given in Hartshorne II.5.18 where $A = k[x,x^{-1}]$, the linear automorphisms are $\psi_1^{-1} \circ \psi_0$ where $\psi_i: \mathcal{O}_{U_i}^r \rightarrow \left.\mathcal{E}\right|_{U_i}$ are isomorphisms, and the definition of isomorphism of vector bundles allows us to change bases of $\mathcal{O}_{U_i}^r$. 
Thinking of this in linear algebra terms, these linear automorphisms on $\mathcal{O}_{U_0 \cap U_1}^r$ are elements of $GL_r(k[x,x^{-1}])$, and changing coordinates in $\mathcal{O}_{U_i}^r$ are elements of $GL_r(k[x])$ for $i = 0$ and $GL_r(k[x^{-1}])$ for $i = 1$. Thus up to isomorphism, the vector bundles of rank $r$ on $\mathbb{P}^1$ are elements of the double quotient $$ GL_r(k[x^{-1}]) \left\backslash \large{GL_r(k[x,x^{-1}])} \right/ GL_r(k[x]).$$ The decomposition of vector bundles into line bundles SHOULD mean that these double cosets can be represented by matrices of the form $$\left(\begin{array}{cccc} x^{n_1} &  &  & 0 \\ & x^{n_2}&  &  \\ &  & \ddots &  \\ 0 &  &  & x^{n_r}\end{array}\right) \quad n_1 \leq n_2 \leq ... \leq n_r.$$ I want to know whether there is a way to prove this fact purely via linear algebra (equivalently, if the geometric proof [cf. Lemma 4.4.1 in Le Potier's "Lectures on Vector Bundles"] has a linear algebraic interpretation). 
[Note: For the affine case, taking the double quotient $$GL_n(k[x]) \left \backslash M_{n,m}(k[x]) \right/ GL_m(k[x])$$ gives the classification of vector bundles over $\mathbb{A}^1_k$ (and of course, when replacing $k[x]$ with an arbitrary PID, gives the usual structure theorem of finitely generated modules over PID).]

REPLY [6 votes]: I know this result as Birkhoff factorization. I was assigned it as an exercise in an algebraic geometry class and came up with the following rather tedious proof. Below I work over $\mathbb{C}$ but the proof makes no use of any facts about the base field.

Any element of $\mathbb{C}[t, t^{-1}]$ can be written uniquely in the form $t^k f(t)$ where $f$ is a polynomial with nonzero constant term. We call $k$ the $t$-adic valuation of $t^k f(t)$.
We want to classify the double cosets 
\begin{equation} \text{GL}_n(\mathbb{C}[t]) \backslash \text{GL}_n(\mathbb{C}[t, t^{-1}]) / \text{GL}_n(\mathbb{C}[t^{-1}]). \end{equation}
Observe that starting from a matrix in $\text{GL}_n(\mathbb{C}[t, t^{-1}])$ and performing $\mathbb{C}[t]$-linear row operations resp. $\mathbb{C}[t^{-1}]$-linear column operations does not change the coset to which the matrix belongs. 
Beginning from such a matrix, perform $\mathbb{C}[t]$-linear row operations as follows. First, consider the $\mathbb{C}[t]$-submodule of $\mathbb{C}[t, t^{-1}]$ generated by the elements of the first column. This submodule is $t^k$ times an ideal of $\mathbb{C}[t]$ for some $k$, which is principal, hence it is generated by one element. Using row operations, replace one of the entries of the column by a generator, place it in the top row, and eliminate the rest of the entries in the column. Repeat for each column. 
After performing the above row operations, we obtain an upper triangular matrix in the same coset as the original. Each of its diagonal entries is a unit in $\mathbb{C}[t, t^{-1}]$, hence must be an invertible constant times $t^k$ for some $k \in \mathbb{Z}$ (and by multiplying the rows or columns by suitable constants we can assume WLOG that they have the form $t^k$). 
Now perform the following sequence of row and column operations on each diagonal above the main diagonal (do not move on to a new diagonal until each entry in the current diagonal is zero). Using row and column operations, ensure that each term of the diagonal entry $p(t)$ has the property that if $t^{\ell}$ is the main diagonal entry to the left of it and $t^d$ is the main diagonal entry below it, then $p(t)$ contains no terms of degree greater than or equal to $d$ or less than or equal to $\ell$. We will abbreviate this situation using the $2 \times 2$ matrix
\begin{equation} \left[ \begin{array}{cc} t^{\ell} & p(t) \\ 0 & t^d \end{array} \right] \end{equation}
where $p(t) = \sum_{m=\ell+1}^{d-1} p_m t^m$. Let $k < 0$ be the negative integer such that the greatest power of $t$ dividing $t^k p(t)$ is $t^{\ell}$, and perform the column operation
\begin{equation} \left[ \begin{array}{cc} t^{\ell} - t^k p(t) & p(t) \\ -t^{d+k} & t^d \end{array} \right]. \end{equation}
Now perform row operations to eliminate the bottom left entry. The result will have the form
\begin{equation} \left[ \begin{array}{cc} t^{\ell'} & p'(t) \\ 0 & t^{d'} \end{array} \right] \end{equation}
where $\ell'$ is the minimum of the $t$-adic valuation of $t^{\ell} - t^k p(t)$ (which is greater than $\ell$) and $d+k$. But since $p(t)$ by assumption contains no terms of degree greater than or equal to $m$, $d+k$ is necessarily also greater than $\ell$. Hence our row and column operations have ensured that $\ell' > \ell$ and $d' < d$. If $p'(t) = 0$, we are done and can move on to a new entry in the same diagonal or to a new diagonal. Otherwise, we repeat. After a finite number of steps, we will have $\ell^{(N)} \ge d^{(N)}$ (where $N$ is the number of steps), at which point $p^{(N)}(t)$ necessarily vanishes and we can move on to a new entry or a new diagonal. 
The above algorithm eventually produces a diagonal matrix with entries of the form $t^k$ for some $k$, which shows that every double coset in
\begin{equation} \text{GL}_n(\mathbb{C}[t]) \backslash \text{GL}_n(\mathbb{C}[t, t^{-1}]) / \text{GL}_n(\mathbb{C}[t^{-1}]). \end{equation}
contains a matrix of the desired form. It remains to be shown that this matrix is unique up to permutation of its diagonal entries. So suppose two diagonal matrices $D, D'$ with diagonal entries $t^{k_i}, t^{k_i'}$ lie in the same double coset, hence there exist $A \in \text{GL}_n(\mathbb{C}[t])$ and $B \in \text{GL}_n(\mathbb{C}[t^{-1}])$ such that
\begin{equation} D' = A^{-1} DB. \end{equation}
By taking determinants we see that $\sum k_i = \sum k_i'$. Rewrite the above identity as
\begin{equation} AD' = DB. \end{equation}
Let $a_{ij}, b_{ij}$ be the entries of $A, B$. Then the above gives
\begin{equation} t^{k_j} a_{ij} = t^{k_i'} b_{ij} \end{equation}
or
\begin{equation} b_{ij} = t^{k_j - k_i'} a_{ij}. \end{equation}
Since $\sum k_i = \sum k_i'$, it follows that $\sum (k_i - k_i') = 0$. If $\sigma \in S_n$ is a permutation, it follows that 
\begin{equation} \sum_i (k_{\sigma(i)} - k_i') = 0 \end{equation}
hence that either $k_{\sigma(i)} = k_i'$ for all $i$ (in which case we are done) or that there exists an $i$ such that $k_{\sigma(i)} - k_i' > 0$. But $a_{ij} \in \mathbb{C}[t]$ and $b_{ij} \in \mathbb{C}[t^{-1}]$, so this is possible if and only if $a_{ij} = b_{ij} = 0$. Since $A, B$ are not identically zero, there must exist a permutation $\sigma$ such that $k_{\sigma(i)} = k_i'$ for all $i$, and the conclusion follows.