# Computing the DBI action on $S^5$

I've been looking at this paper (arXiv: 1103.4079). On page 7, from the metric of the giant gravtiton on $$AdS_5 \times S^5$$,

$$ds^2 = -\cosh^2\rho \, dt^2 + d\rho^2 + \sinh^2 \rho \, d\tilde \Omega_3^2 + d\theta^2 + \sin^2\theta \, d\phi^2 + \cos^2\theta \, d\Omega_3^2,$$

the ansatz

$$\rho=0, \quad \sigma^0=t, \quad \phi=\phi(t) \quad \sigma^i=\chi_i,$$

and the action $$S_{D3} = \frac{-N}{2\pi^2} \int d^4\sigma \; \left( \sqrt{-g} -P[C_4]\right),$$

(where $$g_{ab} = \partial_a X^M \partial_b X_N$$ with $$X^M$$ the embedding coordinates, and $$a,b = 0, \ldots, 3$$) the following formula is given:

$$S = \int dt \; L = -N \int dt\; \left(\cos^3\theta \sqrt{1 - \dot \phi^2 \sin^2\phi} - \dot \phi \cos^4\theta \right)$$

Where does the final line come from? I understand that the final term comes from the pullback in the Wess-Zumino part (the $$P[C_4]$$), and the square root from the determinant of the metric $$g_{ab}$$, but how do I go about explicitly computing it? The embedding metric given lowers the index such that $$g_{ab} = \partial_a X^M \partial_b (G_{MN} X^N)$$, but then I get stuck. How do the derivatives act on the omegas, for example?

It should work as follows. By your "ansatz" the relevant metric reduces to $$ds^2 = -(1- \dot{\phi}^2\sin^2 \theta)dt^2 + d\theta^2 + \cos^2 \theta d\Omega_3^2 \ ,$$ where $$d\Omega_3^2$$ is the line-element of $$S^3$$. The volume of the unit $$S^3$$ is $$\text{Vol}(S^3)=2\pi^2$$.
Now we can calculate the term $$\sqrt{-g}$$. We have (for fixed $$\theta$$) $$\int dt d\Omega_3 \sqrt{-g}=\int dt \sqrt{(1- \dot{\phi}^2\sin^2 \theta) \cdot (\cos^2 \theta)^3} \cdot \text{Vol}(S^3) = 2\pi^2 \int dt \cos^3 \theta \sqrt{1- \dot{\phi}^2\sin^2 \theta}$$
Now we need to calculate $$\int C$$. We are given $$C = C_{\phi \chi_1 \chi_2 \chi_3}d\phi \wedge d\Omega_3 = \dot{\phi}C_{\phi \chi_1 \chi_2 \chi_3}dt \wedge d\Omega_3$$ with $$C_{\phi \chi_1 \chi_2 \chi_3}= \cos^4 \theta \cdot \text{Vol}(S^3)$$. Hence, $$\int C = 2\pi^2 \int \dot{\phi} \cos^4 \theta dt \ ,$$ where we only integrated over the angles of $$S^3$$.
Combining both results yields the asserted equation in the paper. Even the $$2\pi^2$$ cancels out.:-)
• Thank you very much for the answer, just one quick question: where exactly does the $(\cos^2 \theta)^3$ come from? Mar 14 at 19:07
• The radius of the $S^3$ here is $r=\cos \theta$, because we have $ds^2 = ... \cos^2 \theta d \Omega_3^2$. The volume is then $2\pi^2r^3 = 2\pi \cos^3 \theta$. More mathematically, $g$ is the determinant of the metric. Since we have three coordinates $\chi_1,\chi_2,\chi_3$ on $S^3$, we have $\cos^2 \theta$ on the last three diagonal entries of the metric. This results into $(\cos^2\theta)^3$ under the square root.