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  What does the equation $\tau \tau^* = \sigma^* \sigma$ represent in the ADHM construction of vector bundles?

+ 1 like - 0 dislike

I'm looking at the explicit construction of vector bundles with Anti-Self-Dual (ASD) connections on them via the ADHM construction of instantons. At the heart of this is the complex

$$ V \stackrel{\sigma_z}{\rightarrow}V \oplus V \oplus W \stackrel{\tau_z}{\rightarrow}V. $$

Where $\sigma_z = \begin{pmatrix}B_1 - z_1\\B_2 - z_2\\j\end{pmatrix}$ and $\tau_z = \left(-(B_2 - z_2),\; B_1 - z_1,\; I \;\right)$. I am just getting these from Nakajima's book, "Lectures on Hilbert Schemes of Points on Surfaces". And I'll omit explaining the left over notation. Now two key conditions that I find are that $\tau_z \sigma_z = 0$ and $\tau_z \tau^*_z = \sigma^*_z \sigma_z$. The first condition says this is exact, which allows you to look at the quotient $E_z := \text{ker}(\tau_z)/\text{im}(\sigma_z)$. We also require $\sigma_z$ to be injective and $\tau_z$ to be surjective (which is a given assuming ADHM conditions) to assume that $E_z$ never changes rank and in turn will give a vector bundle. So what does $\tau_z \tau^*_z = \sigma^*_z \sigma_z$ tell you? I know that it gives you part of the ADHM condition, but as far as the vector bundle, what does it tell you? I'm pretty sure it gives the Anti-Self-Duality of the induced connection but I'm not sure exactly how to see this. Thanks for the help!!

This post imported from StackExchange MathOverflow at 2015-02-17 11:30 (UTC), posted by SE-user user46348
asked Jan 23, 2015 in Mathematics by user46348 (20 points) [ no revision ]
retagged Feb 17, 2015

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