Let's $\mathbb{S}^d$ be the unit sphere with it's standard metric $g$. A vector field $X \in \mathfrak{X}(\mathbb{S}^d)$ is conformal if and only if there is a function $f \in C^{\infty}(\mathbb{S}^{d})$ such that: \begin{align} \mathcal{L}_Xg=fg \end{align} Killing field are conformal with $f=0$. Taking $a \in \mathbb{R}^{d+1}$, $X(x)=aa\cdot x x$ is conformal too. My question is : is there other examples of conformal field on $\mathbb{S}^d$ which are not combinations of those I gave ?

$\begingroup$ Take a look at mathoverflow.net/questions/60687/theconformalgroupofsn from which the result is not difficult. $\endgroup$ – Ben McKay May 18 '16 at 18:32
The group of conformal diffeomorphisms of the sphere $\mathbb{S}^d$ is isomorphic to SO(d+1,1) via the isomorphism $$(\mathbb{H}^{d+1},\partial \mathbb{H}^{d+1}) \simeq (B^{d+1},\mathbb{S}^d)$$ where $\mathbb{H}^{d+1} = \{(x_0,\ldots,x_{d+1})\in \mathbb{R}^{d+2} x_0^2+x_1^2+\ldots+x_{d+1}^2=1\}$ and $B^{d+1}$ is the $(d+1)$dimensional unit ball with the Poincaré hyperbolic metric $ds=\frac{2dx}{1x^2}$ and where $SO(d+1,1)$ acts on the ambient $\mathbb{R}^{d+2}$ (interpreted as the Minkowski space $\mathbb{R}^{d+1,1}$) leaving $\mathbb{H}^{d+1}$ invariant. All conformal diffeomorphisms of the boundary $\partial \mathbb{H}^{d+1} \simeq \mathbb{S}^d$ extend indeed to the interior $\mathbb{H}^{d+1}$ this way.
All this tells us that the group of conformal diffeomorpshisms of the sphere $\mathbb{S}^d$ has dimension $\dim SO(d+1,1) = \frac{(d+2)(d+1)}{2}$ which is the same as the number of independent conformal vector fields (that generate its Lie algebra).
Isometries, on the other hand, form the group $O(d+1)$ of linear orthogonal transformations of the ambient $\mathbb{R}^{d+1}$ of which $\mathbb{S}^{d}$ is the unit sphere. The number of Killing vector fields is hence $\frac{(d+1)d}{2}$
The difference in dimension between these two groups is precisely $d+1$. Now your vector fields $X_a(x):=a(a\cdot x)x$, $a\in \mathbb{R}^{d+1}$ come as a $(d+1)$dimensional vector space, so the question here is: are any of them actually isometries? Again the answer is no (it is interesting to check it). So you have found all conformal vector fields as either Killing or of the form $X_a(x)$: $$\mathfrak{conf}(\mathbb{S}^d) = \mathfrak{isom}(\mathbb{S}^d) \oplus \{X_a  a\in \mathbb{R}^{d+1}\} $$

1$\begingroup$ In the standard Lie theoretical notation, the isometry group is called $K$ and it is a maxiamal compact in the conformal group $G$. The vector field defined by the OP is the $K$conjugate of the $A$ vector field. The equation by the end of the answer follows from the $KAK$ decomposition. $\endgroup$ – Uri Bader May 18 '16 at 21:02

$\begingroup$ Hi @issoroloap, can you please tell me a reference where I can read more about this? (nice answer BTW, +1). For example, I was under the impression that this equivalence fails for $d=2$? I'd like to see a more detailed discussion. If you know a nice reference, please let me know. Cheers! $\endgroup$ – AccidentalFourierTransform May 15 '17 at 15:39