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Recently I am reading Wall's paper "On the Orthogonal Groups of Unimodular Quadratic Forms II". In this paper, I encountered with the map $E^1_\omega$, which now I am interested in.

Let $X$ be an unimodular integral lattice, and $H$ denotes the unimodular lattice which has a basis $x$, $y$ satisfying $x\cdot x=y\cdot y=0$, and $x\cdot y=1$. For $\omega \in X$ with $\omega\cdot\omega\in 2\mathbb{Z}$, we define the isometry $E^1_\omega$ of $X\oplus H$ as follows.

  • For $\xi\in X$, $E^1_\omega(\xi)=\xi-(\xi\cdot\omega)x$.
  • $E^1_\omega(x)=x$.
  • $E^1_\omega(y)=y+\omega-2^{-1}(\omega\cdot\omega)x$.

My question is: What is algebraic intuition behind the definition of $E^1_\omega$?


EDIT 1 : As D. Ruberman answered below, there exists intuition behind the map $E^1_\omega$ from the view point of differential topology. In fact, Wall proved that, for any closed simply-connected 4-manifold $M$ which has indefinite unimodular form $X$ as intersection form, any automorphism of $X\oplus H$ is realized by some diffeomorphisms of $M\#S^2\times S^2$.

He proved it as follows.

Step 1 : He had showed that the automorphism group of $X\oplus H$ is generated by $E^1_\omega$ and $E^2_\omega$, where $E^2_\omega$ is defined by interchanging the roles of $x$ and $y$ in $E^1_\omega$.

Step 2 : He also proved that each of $E^i_\omega$ is induced from a diffeomorphism of $M\#S^2\times S^2$. This completes the proof.

I already knew the proof of Step 2 which intuitively explains the map $E^i_\omega$ from the topological point of view. But I want to know some intuition behind Step 1, which is algebraic part of the proof.

EDIT 2 : The map $E^i_\omega$ is called the Siegel-Eichler transformation in a lot of literature.

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  • $\begingroup$ Writing $X\oplus H= \langle x \rangle \oplus H \oplus \langle y \rangle$, the given automorphisms are the upper/lower triangular matrices with 1 on the diagonal. $\endgroup$ – user83633 Jul 22 '16 at 12:22
  • $\begingroup$ @user83633 Thank you for your comment. (I think you made a subtle typo $X\oplus H = \langle x \rangle \oplus X \oplus \langle y \rangle$.) $\endgroup$ – Shinichiro Nakamura Jul 24 '16 at 8:31
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This is explained very clearly in Wall's paper, Diffeomorphisms of 4-manifolds, J. London Math. Soc. (1964) 131-140. The idea is that if you do surgery on a circle $C$ in a simply-connected $4$-manifold $X$, you get $X \# T_r$ where $T$ is a 2-sphere bundle over $S^2$ with fiber $S^2$, and intersection form generated by $x,y$ as above, except that maybe $y\cdot y = r$ may be non-zero. (Up to diffeomorphism, only the parity of $r$ matters.) Now if you perform an isotopy of that $C$, ending up at $C$, you get a self-diffeomorphism of $X \# T_r$. (You can get $r\neq 0$ by twisting the framing of the circle in this isotopy.) If the $2$-dimensional homology class swept out by the circle (ie a map of a torus) in this isotopy is $\omega$, then the induced map on homology is of the form $E^1_\omega$.

I don't know what was the original intuition behind the algebraic formula (due to Siegel), but Wall's geometric picture makes it a lot clearer to me.

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  • $\begingroup$ @Ruberman Thank you for your answer. Yes, I have already known geometric intuition (via Kirby calculus) of the map $E^1_\omega$. But I know next to nothing about algebraic point of view, too. $\endgroup$ – Shinichiro Nakamura Jul 22 '16 at 5:57
  • $\begingroup$ Maybe you want to revise your question to ask for the algebraic intuition behind Siegel's definition. $\endgroup$ – Danny Ruberman Jul 22 '16 at 13:36
  • $\begingroup$ @Ruberman You are right. I have edited the question to make it clear. I apologize any confusion because of my bad English language. $\endgroup$ – Shinichiro Nakamura Jul 23 '16 at 4:32
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Recently, I had found some algebraic intuition by generalizing the construction.

We can define similar isomorphism $E^1_a$ of $X\oplus H$ for any $a\in X^{*}(=\text{Hom}_{R}(X,R)$, where $R$ is the coefficient ring of lattices).

To begin with, we define the isomorphism $E^1_a$ of $X\oplus\langle x\rangle$ as follows:

  • For $\xi\in X$, $E^1_a(\xi):=\xi+a(\xi)x$,
  • $E^1_a(x)=x$.

It seems that the construction of $E^1_a$ is natural and easy to come up with. And more, the following lemma holds if we assume the coefficient ring of lattices $R$ has 2 as an unit:

The isomorphism $E^1_a$ can be uniquely extended as isomorphism of $X\oplus H$.

Proof : Let's assume that there exists such an extension and write $E^1_a(y)=\omega+sx+ty$ by using some $\omega\in X$. Because $E^1_a$ preserves the metric, we have

  • For $\xi\in X$, $0=E^1_a(\xi)\cdot E^1_a(y)=\xi\cdot\omega+ta(\xi)$,
  • $1=E^1_a(x)\cdot E^1_a(y)=t$,
  • $0=E^1_a(y)\cdot E^1_a(y)=\omega\cdot\omega +2st$.

Hence, we have $t=1$, for any $\xi\in X$, $\underline{\xi\cdot\omega+a(\xi)=0}_{(*)}$, and $s=-2^{-1}(\omega\cdot\omega)$. Note that there always exists the unique element $\omega\in X$ which satisfies $(*)$ (because $X$ has non-degenerate metric). This suggests the method of construction and completes the proof.

Finally, in the case of $R=\mathbb{Z}$ and $\omega\in X$ with $\omega\cdot\omega\in 2\mathbb{Z}$, if we take $a\in X^*$ as $(*)$, $E^1_a$ agrees with the Siegel-Eichler transformation $E^1_\omega$.

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