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?

  • $\begingroup$ On a sphere $\partial_\theta$ is not an isometry. Use Killing equation for the metric at hand to find isometry vectors $\endgroup$
    – nwolijin
    Jan 29, 2021 at 10:33

1 Answer 1


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.


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