Question on $E_8$ and twistor space

Question

The Kahler $4$ form constructed from two-forms $\{\alpha, \beta\} \in H^2(M,\mathbb Z)$, and $M$ a $4$-manifold, is induced by $\alpha\wedge\beta$ with the map $H^2(M, \mathbb Z)\otimes H^2(M, \mathbb Z) \rightarrow$ $H^4(M, \mathbb Z)$. This defines the topological charge

$$8\pi k = \langle\omega(\alpha\cup\beta)\rangle = \int\omega.$$ This is geometrically the number of ways these two forms intersect, and the four-form $\omega$ is then an intersection form.

Milnor [On Simply Connected 4-manifolds, Symp. Int. Top. Alg., Mexico (1958) 122-128] demonstrated that for

$$M = \{z_0, z_1, z_2, z_4 \in \mathbb CP^3: z_0^4 + z_1^4 + z_2^4 + z_4^4 = 0\},$$ the Kummer surface, that this intersection form is given by $$[E_8]\oplus[E_8]\oplus3\left(\begin{array}{c,c} 0 & 1 \\ 1 & 0\end{array}\right).$$ Here $[E_8]$ is the Cartan center matrix for the exceptional $E_8$ group.

This leads to a couple of questions or observations. One of them is that $\mathbb CP^3$ is projective twistor space $\mathbb P\mathbb T^+$. Twistor projective space is

$$\mathbb P\mathbb T^+ = SU(2,2)/SU(2,1)\times U(1) \simeq SO(4,2)/SO(4,1)\times SO(2).$$ This is a Hermitian symmetric space. The question is then whether the global symmetries given by the Kahler form are a set of global symmetries reduced from the local symmetries of $E_8\times E_8 \sim$ $SO(32)$. This could also be carried to Witten's supertwistor space $\mathbb C\mathbb T^{3|4}$ as well.

String theory requires a background for gravitation. String theory then demands there be some global spacetime symmetry, say a set of global symmetries at the conformal $i^0$. The rest of string theory involves entirely local symmetries. Along the lines of this question, is this relationship between global and local symmetries (assuming my hypothesis here is correct) the reason string theory requires background dependency, such as type II strings on $AdS_5$. The twistor space here has a relationship with the anti-de Sitter spacetime, being quotient groups of $SO(4,2)$ with different divisors.

This post imported from StackExchange Physics at 2016-10-02 10:56 (UTC), posted by SE-user Lawrence B. Crowell

Comments

Your introductory paragraph does not make sense to me. A Kähler form is conventially the symplectic form of a Kähler manifold, there is no such thing as a "Kähler form of two 2-forms". The equality $H^4(M,\mathbb{Z}) = H^2(M,\mathbb{Z})\otimes H^2(M,\mathbb{Z})$ is just wrong for a general 4-manifold, just take $S^4$ where the l.h.s. is $\mathbb{Z}$ and the r.h.s. $0$ to see this. What is $\alpha\cup\beta$ and where does $\omega$ suddenly come from? Is it supposed to be a fundamental class? The "intersection form" is the map $(\alpha,\beta)\mapsto\int_M \alpha\wedge\beta$, not some diff-form

This post imported from StackExchange Physics at 2016-10-02 10:56 (UTC), posted by SE-user ACuriousMind

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Also, why are there "global symmetries given by the Kähler form"? What does that mean? What is the expression $E_8\times E_8\sim \mathrm{SO}(32)$ supposed to denote? The two groups have the same dimension, but are not the same in any way!

This post imported from StackExchange Physics at 2016-10-02 10:56 (UTC), posted by SE-user ACuriousMind

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I changed the wording, that was admittedly awkward. The equality is supposed to be a map induced by the joining or union. The heterotic string theory quite frequently makes use of the similar Dynkin diagrams and the same number of roots between $E_8\times E_8$ and $SO(32)$. This correspondence is to my knowledge not fully understood.

This post imported from StackExchange Physics at 2016-10-02 10:56 (UTC), posted by SE-user Lawrence B. Crowell

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Lawrence B. Crowell, John Rennie and @Danu: Actually, moderators can no longer merge accounts. The user should instead go here and follow the instructions. If all that fails, the user should contact the SE team.

This post imported from StackExchange Physics at 2016-10-02 10:56 (UTC), posted by SE-user Qmechanic