# Question about Type IIB supergravity equations of motion

This is probably a dumb question, but I'm a mathematician who's been trying to understand the equations of motion for Type IIB supergravity, and I'm not quite sure I understand what's going on with the Einstein equations. Specifically, I'm following the appendix A of this paper, and the equation I'm having trouble with is A.4. Here it is for clarity:

$$R_{\mu \nu} + 2 D_{\mu}D_{\nu}\Phi - \frac{1}{4}H_{\mu \nu}^2 = e^{2\Phi}\left[ \frac{1}{2}(F_1^2)_{\mu \nu} + \frac{1}{4}(F_3^2)_{\mu \nu} + \frac{1}{96}(F_5^2)_{\mu \nu} - \frac{1}{4}g_{\mu \nu}\left( F_1^2 + \frac{1}{6}F_3^2 \right )\right]$$

My understanding is that $$F_1$$ is a differential 1-form, which we would write in components as say $$F_1 = f_i dx^i$$, and that $$F_1^2$$ is defined then to be: $$F_1^2 = f_i f_j g^{ij}$$ My problem is therefore that I expect $$F_1^2$$ to be a scalar. Indeed, I expect all the terms on the RHS of the equation to be scalars, and so I don't know what $$(F_1^2)_{\mu \nu}$$ means.

Small addendum: I am fairly sure that $$D_{\mu }$$ is the covariant derivative with respect to the metric connection, something I would normally call $$\nabla_{\mu}$$, but I just wanted to confirm this.

For a target space $p$-form in local coordinates

$$F_p~=~\frac{1}{p!}F_{\mu_1\mu_2\ldots\mu_p}~\mathrm{d}x^{\mu_1}\wedge \mathrm{d}x^{\mu_2}\wedge\ldots\wedge \mathrm{d}x^{\mu_p},\tag{1}$$

the paper defines a scalar

$$F_p^2~:=~F_{\mu_1\mu_2\ldots\mu_p}~g^{\mu_1\nu_1}~g^{\mu_2\nu_2}\ldots g^{\mu_p\nu_p}~F_{\nu_1\nu_2\ldots\nu_p},\tag{2}$$

and a symmetric covariant tensor

$$(F_p^2)_{\mu_1\nu_1}~:=~F_{\mu_1\mu_2\ldots\mu_p}~g^{\mu_2\nu_2}\ldots g^{\mu_p\nu_p}~F_{\nu_1\nu_2\ldots\nu_p}.\tag{3}$$

$F_{n-1}$'s are curvatures of higher gauge fields $A_{n-1}$'s, so they are n-forms. The expression $(F_{n}^2)_{\mu \nu}$ means schematically $F_{\mu \rho_1 \cdots \rho_{n-1}} F_{\nu}^{\phantom{a} \rho_1 \cdots \rho_{n-1}}$, where the indices are risen with the inverse metric $g^{\mu \nu}$.

• Does this mean that $F_1^2$ is just $F_{\mu} F_{\nu}$? – Mark B Feb 11 '17 at 7:16
• @MarkB Sorry, I made a typo in my answer. $F_1$ is the curvature of $A_1$, $F_1=\mathrm{d} A_1$. – Andrey Feldman Feb 11 '17 at 7:23
• Yep, I'm ok with the fact that $F_1$ is the curvature of a scalar field, $F_1 = d C_0$. But a one-form has only one index, so according to your answer the square of the 1-form flux is simply $F_{\mu} F_{\nu}$, which seems a bit strange to me. – Mark B Feb 11 '17 at 7:27
• @MarkB Sorry, you are right. There are, of course, no odd-degree gauge fields in Type IIB. The correct expressions are given by (2.4) and (2.5) in the paper you cited. Then, $F_1^2$ is just $\partial_{\mu} C_0 \partial_{\nu} C_0$. – Andrey Feldman Feb 11 '17 at 7:34