Linear Algebra – Using Linear Algebra to Classify Vector Bundles

ag.algebraic-geometrylinear algebravector-bundles

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).]

Best Answer

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.

Related Question