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I'm studying an example provided by Hatcher in his K-theory and Vector Bundle book. I'm referring to example 1.13 pag. 24

example 1.13

The first part is clear, $z^2$ is the Kronecker Product of the two clutching function (or transition maps) for the taut-line-bundle, but then I can't follow is reasoning about the generalization.

1) Is $H \otimes H$ a special case of pullback of $H$?

2) Why is the matrix-representation (we fixed a point of $S^{k-1}$) for the pointwise product of clutching functions the Schur Product (or Hadamard or Pointwise product) of the two matrices?

3) Does $\left(f\oplus \text{Id} \right)\alpha_t\left(\text{Id} \oplus g \right)\alpha_t$ should be $\left(f\oplus \text{Id} \right)\left(\text{Id} \oplus g \right)\alpha_t$ ? I'm not sure, because if $t=1$ (or $t=0$) we have $\left(f\oplus \text{Id} \right)\left(\text{Id} \oplus g \right)$ which is $\left( f \oplus g \right)$, but for $t=0$ (or $t=1$) we have $\left(f\oplus \text{Id} \right)\left( g \oplus \text{Id} \right)$ but this is NOT equal to $\left( fg \oplus \text{Id} \right)$ because the matrix of $fg$ is the Schur product, not the usual matrix product). The problem is I don't understand the meaning of the $\alpha_t$ in the middle.

Riccardo
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  • http://math.stackexchange.com/questions/41936/understanding-the-canonical-line-bundle-h-and-the-fact-that-h-otimes-h-op?rq=1 – cactus314 Jul 05 '14 at 15:55
  • @johnmangual well, the question you linked is about the first half of the example, which gave me no problem, I'm asking for the second half. Thanks anyway:) – Riccardo Jul 05 '14 at 15:57

1 Answers1

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1) it is $H=E_z$ and so $H\otimes H=E_{z^2}$.

2) I think it is the usual Matrix product. The pointwise refers to points on $S^{k-1}$. For every point $x\in S^{k-1}$ you obtain two matrices by applying f and g. The product clutching function evaluated at any point should be the product of these two.

3) works fine as well if you take the usual matrix product

jorst
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  • Agree with 1) I don't mean to be rude, I have the same feeling about 2) but I need some kind of stronger argument to convince myself rather than "it should be" :(, maybe an example. so in 3) is there an error as I pointed out? – Riccardo Jul 05 '14 at 15:21
  • Obviously, thanks for the answer anyway ! – Riccardo Jul 05 '14 at 15:22
  • there is no error in 3) (if you buy that you have to use normal matrix multiplication) because you have $\alpha_t^{-1}=\alpha_t$ and so you conjugate the matrix – jorst Jul 05 '14 at 15:32
  • the matrix $\left( fg \oplus \text{Id} \right)$ is in fact the usual matrix product of $\left(f\oplus \text{Id} \right)\left( g \oplus \text{Id} \right)$ (direct sum of two matrices gives you a blog matrix with the one on the top left and the other one on the bottom right and so multiplication can be performed blockwise) – jorst Jul 05 '14 at 15:35
  • Would you mind elaborate a little more why is the conjugate involved? It seems that I miss the point :( – Riccardo Jul 05 '14 at 15:47
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    You want to exchange the two blocks in the matrix. Therefore you write it in a different Basis (namely the one with the first n coordinates interchanged with the last n). Change of basis is done by conjugation by a matrix that sends your previus basis to the new one. – jorst Jul 05 '14 at 16:05