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

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).]

share|improve this question
add comment

4 Answers

up vote 9 down vote accepted

I must admit I have never read this reference, but I remember it from a similar discussion on a German forum, according to which there is a simple proof in

Michiel Hazewinkel and Clyde Martin, A short elementary proof of Grothendieck's theorem on algebraic vectorbundles over the projective line, Journal of pure and applied algebra 25 (1982), pp. 207 - 211.

share|improve this answer
add comment

Dear Ila, the linear algebra result you mention is due to Dedekind-Weber and was published in Crelle's Journal dated 1882, in their article "Theorie der algebraischen Funktionen einer Veränderlichen". Their motivation was proving Riemann-Roch on an arbitrary smooth projective curve $X$ by presenting the curve as a ramified covering of $\mathbb P^1$ and pushing down the line bundle associated to a divisor on $X$ (they used the language of function fields).

Of course Grothendieck was not aware of this result, which was also rediscoverd by Birkhoff in an analytic setting in 1913, by Plemelj in1908, by Hilbert in 1905...

You will find some details on the metamorphosis of this linear algebra theorem into Grothendieck's result in Scharlau's interesting paper


share|improve this answer
This article is really interesting! Thanks. –  Ila Varma Feb 26 '10 at 17:48
add comment

When I was a graduate student, my advisor Phillip Griffiths told me that the Grothendieck splitting theorem was equivalent to the Kronecker pencil lemma, which gives a normal form for a 2-dimensional space of rectangular matrices. I recall working out the equivalence, and I actually used it in my Ph.D. thesis (which was not on algebraic geometry). The statement and proof of the Kronecker pencil lemma can be found in Gantmacher's book, "The Theory of Matrices" and relies only on linear algebra. I don't know anything about the Dedekind-Weber result cited by Georges Elencwajg. I recall that I found the book "Vector Bundles on Complex Projective Spaces" by Okonek et al to be very helpful.

share|improve this answer
add comment

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}


\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.

share|improve this answer
"Using row operations, replace one of the entries of the column by a generator" -- How does this work? Assuming that the column consists of elements $u_1,u_2,...,u_n$, and that the ideal generated by these elements is also generated by their linear combination $\lambda_1 u_1 + \lambda_2 u_2 + ... + \lambda_n u_n$ alone (with $\lambda_i \in \mathbb{C}\left[t\right]$), I could of course replace (say) $u_1$ by $u_1 + \lambda_2 u_2 + ... + \lambda_n u_n$. But I couldn't multiply $u_1$ with $\lambda_1$. I understand that this would work if $\mathbb{C}\left[t\right]$ was a local ring... –  darij grinberg Sep 23 '13 at 22:57
Oh, I see. You are writing $u_j = t^k v_j$ with $v_j \in \mathbb{C}\left[t\right]$, and then you are reducing the $v_j$ modulo each other as in the Euclidean algorithm, making the appropriate row operations in each step. This works because $\mathbb{C}\left[t\right]$ is an Euclidean domain. Nice! –  darij grinberg Sep 23 '13 at 23:00
"Repeat for each column" -- not literally. Rather, in order not to unsettle the already cleared-out columns, we must repeat the row operations for the entries $2$ to $n$ of the $2$nd column, then for the entries $3$ to $n$ of the $3$rd column, then for the entries $4$ to $n$ of the $4$th column etc.. –  darij grinberg Sep 23 '13 at 23:04
The resulting matrix is upper-triangular. The fact that each of its diagonal entries is a unit in $\mathbb{C}\left[t,t^{-1}\right]$ follows from the fact that the product of these entries is the determinant, which is unchanged by all the row operations and thus invertible (since the original matrix was in $\mathrm{GL}_n\left(\mathbb C\left[t,t^{-1}\right]\right)$). –  darij grinberg Sep 23 '13 at 23:05
I'm stuck at "and perform the column operation". Assume that we are clearing out the diagonal which contains $p(t)$ from northwest to southeast. Can't the column operation "dirty" a cell that was already cleared out? And if we're doing it the other way round, what about the row operation? –  darij grinberg Sep 23 '13 at 23:16
show 4 more comments

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.