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I am trying to understand Example 1.13 of Hatcher's book on vector bundles and K-Theory (page 24).

The cannonical line bundle $H \to \mathbb{C}P^1$ satisfies the relation $(H \otimes H)\oplus 1 \simeq H \oplus H$.

The total space, $H$, is given by $$H = ((x,v) \in \mathbb{C} P^1 \times \mathbb{C}^2:v \in X \},$$ with the projection map $(x,v) \mapsto x$

To prove this, I need to understand clutching functions. The following is from Atiyah's book 'K-Theory':

Suppose we have a space $X = X_1 \cup X_2$ such that $X_1 \cap X_2 = A$. Assume also we have vector bundles $p_i:E_i \to X_i$ and that $\phi:E_1|A \to E_2|A$ is an isormophism.

Then, we can form a vector bundle $E_1 \cup_\phi E_2 \to X$.

Alternatively, Hatcher defines a map $f:A \to GL_n(\mathbb{C})$ (although he specifically constructs clutching functions from spheres, not arbitrary spaces). The two definitions are equivalent.

In particular we can construct a clutching function over the complex line bundle $\mathbb{C} P^1 \simeq S^2$. This is example 1.10 of Hatcher's book. I am not 100% confident in the argument, but I think I understand it. Regardless, the clutching function derived is $f(z)=(z)$ (i.e. multiplication by $z$)

Now I start to get hazy, in Example 1.13. To quote verbatim:

Let us show that the canonical line bundle $H \to \mathbb{C}P^1$ satisfies the relation $(H \otimes H)\oplus 1 \simeq H \oplus H$ where 1 is the trivial one-dimensional bundle. This can be seen by looking at the clutching functions for these two bundles, which are the maps $S^1 \to GL_2(\mathbb{C})$ given by

$$z \mapsto \begin{pmatrix} z^2 & 0 \\ 0 & 1 \end{pmatrix} $$

and

$$z \mapsto \begin{pmatrix} z & 0 \\ 0 & z \end{pmatrix} $$

I get lost in this last argument. I can buy the clutching function for the identity is the second matrix, but where did the first come from? Why do we work over $GL_2(\mathbb{C})$? (In Example 1.10 we worked over $GL_1(\mathbb{C})$.

Moreover, even if I believe these clutching constructions, how can they be used to show that $(H \otimes H)\oplus 1 \simeq H \oplus H$?

Any advice, or references appreciated

Update: Theo's comments below, made me realise I was way off with my thinking. I think the following makes some sense.

Firstly given we have the clutching function for $(z)$ for $H$, then the clutching function for $H \otimes H$ is $(z^2)$. The clutching function for the trivial bundle is the identity.

I am slighly confused as to why we can then take the direct sum of the bundles as the matrix with components along the diagonal. My best thought is that there is a group homomorphism $GL(m,\mathbb{C}) \times GL(n,\mathbb{C}) \to GL(m+n,\mathbb{C})$ given by $$\Theta(A,B) = \begin{pmatrix} A & 0 \\ 0 & B \end{pmatrix} $$

From this, it is clear how to construct the clutching functions above. Hatcher gives the general homotopy via a path $\alpha_t \in GL_{2n}(\mathbb{C})$ from the identity matrix, to the matrix of the transformation which interchanges the two factors of $\mathbb{C}^n \times \mathbb{C}^n$

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Do you buy that $H \otimes H$ has $(z^2)$ as clutching function? Then simply add $1$ to it to get the first matrix. Moreover, it may be worthwhile to read the example until its end :) –  t.b. May 29 '11 at 16:47
    
@Theo - I do buy that. In fact I was totally off - I thought the first matrix was a clutching function for $H$, not $(H \otimes H) \oplus 1$. I will update the question –  Juan S May 30 '11 at 0:21
    
sorry for commenting on an old post but one way to derive the clutching function for $H$ is to look at at typical element at $|z| =1$. That is: $([z:1], (\lambda z, \lambda)) = ([1: 1/z], (\lambda z, \lambda))$ since $z \not= 0$. We would like the vector entry in the second expression to be $(\lambda, \lambda z^{-1})$ as that is what a typical element is in the bundle over $D_{\infty}$. So you need to multiply by $z^{-1}$! –  Elden Elmanto Oct 1 '13 at 3:46
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up vote 4 down vote accepted

I'll work most of this answer with the example $H^{-1} \oplus H \cong \mathbb{C}^2$, because I find it easier to think about. This means that the clutching functions are $\left( \begin{smallmatrix} z & 0 \\ 0 & z^{-1} \end{smallmatrix} \right)$ and $\left( \begin{smallmatrix} 1 & 0 \\ 0 & 1 \end{smallmatrix} \right)$. At the end, I'll make a minor comment about switching things up to do your example.

This is an interesting example because the above isomorphism is correct as topological, or smooth, vector bundles, but is NOT valid as holomorphic vector bundles. So the isomorphism will have to involve some non-holomorphic maps, even though the two vector bundles are holomorphic.

One quick note: The bundle you call "canonical" is not the one I learned to call canonical. I always called your $H$ the "tautological" line bundle.


As you say above, the fiber of $H$ above $(u:v)$ is $\mathbb{C} \cdot (u,v) \subseteq \mathbb{C}^2$. So we have an obvious inclusion $H \hookrightarrow \mathbb{C}^2$. The line bundle $H^{-1}$ is the dual to $H$. Fix a skew symmetric bilinear form $\omega $ on $\mathbb{C}^2$. Then there is a map $\mathbb{C}^2 \to H^{-1}$ taking the point $(x,y)$ to the linear functional $\omega( (x,y), \bullet)$. The kernel of this map is precisely the image of $H$. That is to say, the linear form $\omega( (x,y), \bullet)$ restricts to $0$ on $\mathbb{C} \cdot (u,v)$ precisely if $(x,y) \in \mathbb{C} \cdot (u,v)$. So we have a short exact sequence $$0 \to H \to \mathbb{C}^2 \to H^{-1} \to 0.$$ To be concrete, I'll take $\omega( (x_1, y_1), (x_2, y_2) ) = x_1 y_2 -x_2 y_1$.

Now, in the smooth category, every short exact sequence splits. So $\mathbb{C}^2 \cong H \oplus H^{-1}$ as claimed. Let's work out the isomorphism explicitly.

Recall the proof that every short exact sequence splits. Fix a positive definite Hermitian form on the middle term: We'll choose $\langle (x_1, y_1), (x_2, y_2) \rangle = x_1 \overline{x_2} + y_1 \overline{y_2}$. We are supposed to be split the sequence by lifting a vector $p$ in $H^{-1}$ to its unique preimage in $\mathbb{C}^2$ which is $\langle, \rangle$-orthogonal to $H$.

Let's suppose we are at the point $(x:y)$ in $\mathbb{P}^1$. Let our vector in $H^{-1}$, above the point $(x:y)$, be the linear functional $(x,y) \mapsto 1$. So the preimage of this point in $\mathbb{C}^2$ is $\{ (x', y') : x' y - y' x= 1\}$. The condition that $(x', y')$ be orthogonal to $H$ is that $x' \overline{x} + y' \overline{y} =0$. Solving these linear equations, we get $$(x', y') = \left( \frac{\overline{y}}{|x|^2+|y|^2}, \frac{- \overline{x}}{|x|^2+|y|^2} \right).$$

So, here is our isomorphism $H \oplus H^{-1} \to \mathbb{C}^2$. I'll describe it in the fiber over $(x:y)$. A basis of the fiber of $H \oplus H^{-1}$ is $e:=(x,y)$, in the first summand, and $f: = \left( (x,y) \mapsto 1 \right)$ in the second summand. Our trivialization sends $e \mapsto (x,y)$ and $f \mapsto (\overline{y}, - \overline{x}) /(|x|^2+|y|^2)$. Exercise: The map $H \oplus H^{-1} \to \mathbb{C}^2$ which I have described is independent of which representative $(x,y)$ we have chosen for $(x:y)$.


I'm running out of time to work on this. So I'll write down what this turns into in terms of clutching functions. There may be some minor errors here. We have $$\begin{pmatrix} z \vphantom{\frac{z}{1+|z|^{-2}}} & 1 \vphantom{\frac{z}{1+|z|^{-2}}} \\ \frac{-1}{1+|z|^2} & \frac{\overline{z}}{1+|z|^2} \end{pmatrix} \begin{pmatrix} 1 \vphantom{\frac{z}{1+|z|^{-2}}} & 0 \vphantom{\frac{z}{1+|z|^{-2}}} \\ 0 \vphantom{\frac{z}{1+|z|^{-2}}} & 1 \vphantom{\frac{z}{1+|z|^{-2}}} \end{pmatrix} \begin{pmatrix} \frac{1}{1+|z|^{-2}} & - z^{-1} \\ \frac{\overline{z}^{-1}}{1+|z|^{-2}} & 1 \end{pmatrix} = \begin{pmatrix} z \vphantom{\frac{z}{1+|z|^{-2}}}& 0 \\ 0 & z^{-1}\vphantom{\frac{z}{1+|z|^{-2}}} \end{pmatrix} $$

Note that the left matrix is smooth and invertible on the chart where $z \neq \infty$, and the right matrix is smooth and invertible on the chart where $z \neq 0$. So this shows that the two clutching functions give isomorphic vector bundles.

To return to your original request to equate $H^{\otimes 2} \oplus \mathbb{C}$ and $H^{\oplus 2}$, take the above equation and multiply both sides by $z$.

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Thanks David! I haven't thought about this for a while, so it might take some time to digest your answer –  Juan S Feb 1 '12 at 6:02
    
what an answer! –  Elden Elmanto Oct 1 '13 at 3:47
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