# Isometries or Isometry direction of $AdS_5 \times S^5$

This is a consequent question with my previous question https://physics.stackexchange.com/q /610501/

I want to know the isometries (or isometry direction) of $$AdS_5 \times S^5$$. Usually, when we consider Schwarzschild Black holes, time and angle are isometry direction.

Simply consider $$2-sphere$$, the metric is given by \begin{align} ds^2 = d\theta^2 + \sin^2(\theta) d\phi^2 \end{align} there are two generators of isometries, $$U = \partial_{\phi}, V = \partial_{\theta}$$.

So suppose I have $$AdS_5 \times S^5$$ in Poincare patch \begin{align} ds^2 = R^2 \left( \frac{dz^2}{z^2} - \frac{dt^2}{z^2} + \frac{d\vec{x}^2}{z^2}+ \cos^2(\theta) d \phi^2 + d\theta^2 + \sin^2(\theta) d \tilde{\Omega}_3^2 \right) \end{align} where $$d\vec{x}^2 = dx_1^2 + dx_2^2 + dx_3^2$$, then what is the isometry directions?

• On a sphere $\partial_\theta$ is not an isometry. Use Killing equation for the metric at hand to find isometry vectors – nwolijin Jan 29 at 10:33

The best way to see the isometries of $$AdS_5\times S^{5}$$ is to consider it as embedded in $$\mathbb{R}^{2,4}\times \mathbb{R}^{6}$$. The embedding is possible and it is given by the simple formulas

$$\eta_{MN}^{(2,4)}X^{M}X^{N}=-R^{2},\qquad \delta_{IJ}Y^{I}Y^{J}=R^{2}$$

where $$M=-1,0,1,2,3$$ are the directions of $$\mathbb{R}^{2,4}$$ and $$I=1,2,3,4,5,6$$ are the directions of $$\mathbb{R}^{6}$$. The $$\eta^{(1,4)}_{MN}$$ is the matrix $$\text{diag}(-1,-1,+1,+1,+1,+1)$$. I think it is pretty obvious that the equation above for $$Y^{I}$$ defines an $$S^{5}$$, so what may be new for you is that the equation for $$X^{M}$$ defines the $$AdS_5$$ space. One way to show that is to solve the constraint and see that the metric $$\eta_{MN}^{(2,4)}$$ inherent from the parent space $$\mathbb{R}^{2,4}$$ is the metric that defines your $$AdS_5$$ space.

The isometries are simply the obvious symmetries of the equation above, that is the $$SO(2,4)\times SO(6)$$ transformations that preserve the $$\eta_{MN}^{(2,4)}$$ and $$\delta_{IJ}$$ flat metrics.

We can also do something even better and recognize that $$AdS_5\times S^{5}$$ is an homogenous space, i.e. it is possible to start in an arbitrary point $$p_0$$ and reach a point $$p$$ just by isometry transformations! This means that if we fix an origin, let us say $$p_0$$, we can view our space as

$$AdS_5\times S^{5}= \frac{SO(2,4)\times SO(6)}{SO(1,4)\times SO(5)}$$

where the quotient is due to the fact that a subgroup of the isometry group preserve the origin $$p_0$$. Explicitly this goes as

$$p= g \star p_0,\qquad p,p_0\in AdS_5\times S^{5},\qquad g\in SO(2,4)\times SO(6)$$

where the subgroup $$SO(1,4)\times SO(5)$$ is determined by the equation

$$p_0=h\star p_0$$

Here, the operation $$\star$$ means that we are acting with the isometry element $$g$$ in $$p_0$$.

The subgroup in the quotient act as a gauge symmetry, i.e. a redundancy, and it must be fixed in order to pass through each point once. This gauge fixing is usually done by the exponential map, i.e. we do something like

$$p(\theta)=\exp( \theta^\alpha T_{\alpha})\star p_0$$

where the generators $$\{T_{\alpha}\}$$, are chosen in order to not generate $$SO(1,4)\times SO(5)$$ gauge orbits, so it is a good exercise to you to prove that $$\alpha=1,\dots ,10$$, i.e. that the number of generators should be ten.

Let us define $$g(\theta)=\exp( \theta^\alpha T_{\alpha})$$, then the Killing vectors $$v^{a}$$ associated to the generator $$T_{a}$$ of $$SO(2,4)\times SO(6)$$ will be given by

$$v^{a}(\theta)T_{a}= \left(g^{-1}(\theta)T_{a}g(\theta)\right)$$

If you want to obtain the Killing vectors of $$AdS_5\times S^{5}$$ in a particular coordinates you can try to view your coordinates as a gauge fixing of $$SO(1,4)\times SO(5)$$ of the coset above.