Abstract
We study gravitational theories with a cosmological constant and the Gauss-Bonnet curvature squared term and analyze the possibility of de Sitter expanding spacetime with a constant internal space. We find that there are two branches of the de Sitter solutions: Both the curvature of the internal space and the cosmological constant are (1) positive and (2) negative. From the stability analysis, we show that the de Sitter solution of the case (1) is unstable, while that in the case (2) is stable. Namely de Sitter solution in the present system is stable if the cosmological constant is negative. We extend our analysis to the gravitational theories with higher-order Lovelock curvature terms. Although the existence and the stability of the de Sitter solutions are very complicated and highly depend on the coupling constants, there exist stable de Sitter solutions similar to the case (2). We also find de Sitter solutions with Hubble scale much smaller than the scale of a cosmological constant, which may explain a discrepancy between an inflation energy scale and the Planck scale.
Highlights
Test branes and do not take into account the back reactions
We find that there are two branches of the de Sitter solutions: both the curvature of the internal space and the cosmological constant are (1) positive and (2) negative
It is remarkable that we have de Sitter solution even for a negative cosmological constant
Summary
We consider the following low-energy effective action for the heterotic string with a cosmological constant Λ:. Let us consider the metric in D-dimensional space, ds2D = −e2u0(t)dt2 + e2u1(t)ds2p + e2u2(t)ds2q ,. The external p-dimensional and internal q-dimensional spaces (ds2p and ds2q) are chosen to be maximally symmetric, with the signature of the curvature given by σp and σq, respectively. RG2 B = e−4u0 p3A2p + 2p1q1ApAq + q3A2q + 4u 1u 2(p2qAp + pq2Aq) + 4p1q1u 21u 22. F2 = α2e−2u0 p3A2p + 2p1q1ApAq + q3A2q + 4(p2qAp + pq2Aq + p1q1u 1u 2)u 1u 2 , f2(p) = α2e−2u0 (p−1)4A2p +2(p−1)2q1ApAq +q3A2q +4 (p−1)3qAp +(p−1)q2Aq +(p−1)2q1u 1u 2 u 1u 2 , f2(q) = α2e−2u0 p3A2p +2p1(q −1)2ApAq +(q −1)4A2q +4 p2(q −1)Ap +p(q −1)3Aq +p1(q −1)2u 1u 2 u 1u 2 , g2(p) = 4(p − 1)α2e−2u0 (p − 2)3Ap + q1Aq + 2(p − 2)qu 1u 2 , g2(q) = 4(q − 1)α2e−2u0 p1Ap + (q − 2)3Aq + 2p(q − 2)u 1u 2 , h(2p) = 4qα2e−2u0 (p − 1)2Ap + (q − 1)2Aq + 2(p − 1)(q − 1)u 1u 2 , h(2q) = 4pα2e−2u0 (p − 1)2Ap + (q − 1)2Aq + 2(p − 1)(q − 1)u 1u 2.
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