Towards Matrix Models in IIB Superstrings

arXiv:hep-th/9708070v1 12 Aug 1997

TOWARDS MATRIX MODELS OF IIB SUPERSTRINGS

P. Olesen The Niels Bohr Institute Blegdamsvej 17 DK-2100 Copenhagen Ø Denmark Invited talk at the “Nato Advanced Research Workshop on Theoretical Physics”, June 14-20, 1997, Zakopane, Poland

INTRODUCTION Recently there has been a lot of papers on matrix models and superstrings, induced by the work of Banks, Fischler, Shenker, and Susskind (hep-th/ 9706168). I refer to Makeenko’s talk at this meeting for a general review of this subject. Most of the work∗ reported in this talk has been done together with Fayyazuddin, Makeenko, Smith, and Zarembo1 . As explained in Makeenko’s (virtual) talk at this meeting, we started from the work by Ishibashi, Kawai, Kitazawa, and Tsuchiya2 , who proposed that type IIB superstrings in 10 dimensions are described by the reduced action,   1 1 2 µ ¯ [Aµ , ψ]) , SIKKT = α − Tr [Aµ , Aν ] − Tr (ψΓ (1) 4 2 where Aµ and ψα are n × n matrices. A sum over n is implied, with weight exp(−βn). Later the sum over n has been replaced by a double scaling limit3 . In our paper1 we discussed various problems associated with eq. (1), and we proposed a different model with action
α α ¯ µ [Aµ , ψ]), SNBI = − Tr Y −1 [Aµ , Aν ]2 + V (Y ) − Tr (ψΓ 4 2

(2)

where the potential is given by

V (Y ) = β TrY + γ Tr ln Y. ∗

(3)

The paper in reference 1 has been published in Nuclear Physics B. Unfortunately, the editors of that journal used an early draft of the manuscript, which contained several typhos. For this reason I cannot recommend the published version, but refer to the version in the Archives.

1

The partition function is thus given by Z Z = dAµ dψdY e−SNBI .

(4)

We select the constant γ in such a way that the result of the Y -integration is as close to the superstring as is possible. This turns out to mean 1 γ =n− , 2

(5)

as we shall see later. Physically the model SNBI is motivated by a GUT scenario: Suppose one has a field theory valid down to the GUT scale. Then, in our model the group is SU(n), with n large. As we shall see, this type of GUT model then leads to superstrings if n → ∞. For n finite, supersymmetry is broken, as is expected for energies below the GUT energy. Thus, superstrings can emerge from a GUT type of model. Of course, the model with action SNBI is not a realistic GUT model. Under the SUSY transformations i ǫΓµ ψ, δψ = {Y −1 , [Aµ , Aν ]}Γµν ǫ, δAµ = i¯ 4

(6)

the action transforms like
δSNBI ∝ ǫµαβλ1 ...λ7 Tr ψm (Γ0 Γ11 Γλ1 ...Γλ7 )mp ǫp {[Aα , Aβ ], [Aµ , Y −1 ]} .

(7)

δSNBI → 0 for n → ∞,

(8)

It can be shown that

so the action is supersymmetric in the limit n → ∞, but for finite n the symmetry is broken. THE Y -INTEGRATION The integration over Y can be done exactly. Consider Z  α  −1 2 F (z) = dY exp − Tr(Y z ) − β TrY − γ Tr ln Y , z 2 ≡ −[Aµ , Aν ]2 . 4

(9)

The “angular” integration is of the Itzykson-Zuber type, so we get F (z) = const.

i=n Z Y i=1

dyi

P P P ∆2 (y) −α i zi2 /4yi −β i yi −γ i ln yi e . ∆(1/y)∆(z 2 )

Here the zi ’s and yi ’s are the eigenvalues, and Y ∆(x) = (xi − xj ) = det xik−1 ki

i>j

(10)

(11)

is the Vandermonde determinant. We only integrate over the positive eigenvalues of Y . Thus we get Z ∞Y Y P 2 P P 2 ∆(z )F (z) = const. dyi yin−1 (yi − yj ) e−α i zi /4yi −β i yi −γ i ln yi . (12) 0

2

i

i>j

This can be rewritten as a determinant Z dy 2 2 ∆(z )F (z) = const. det √ y k−1e−αzi /4y−βy ki y r   k−1 π −√αβzi k−1 ∂ e . = const. det (−1) ki ∂β k−1 β

(13)

Here zi is by definition the positive square root of zi2 . This determinant can be evaluated using basic properties of determinants, and the result is1 ∆(z 2 )F (z) = const. ∆(z) e−

√

αβ

P

i zi

.

(14)

This result is exact, and hence it is valid for any n. The sum over eigenvalues in the exponent has the following interpretation, X

zi =

i

Xq

zi2

i

1 = √ 4i π

Z

(0+)

−∞

dt X tzi2 e , t3/2 i

(15)

where we used an integral representation of the square root. Thus, X i

1 zi = √ 4i π

Z

(0+) −∞

dt Tr exp(−t[Aµ , Aν ]2 ) = Tr 3/2 t

q

−[Aµ , Aν ]2 ,

(16)

valid in Euclidean space. The partition function therefore becomes Z =

α

Z



2

−1

dAµ dψdY exp Tr(Y [Aµ , Aν ] ) − β TrY − γ Tr ln Y 4   Z q p α dAµ dψ µ ¯ [Aµ , ψ]) . exp − αβ Tr −[Aµ , Aν ]2 − Tr (ψΓ = const. Q 2 i>j (zi + zj )

(17)

This is the exact result of the Y -integration. In order to get the square root it is important to use the value of γ given in eq. (5). This result can be expressed in an alternative form, at the cost of introducing an auxillary Hermitean field M. We use the identity √ 1 = const. det z i