# Conformal Killing fields on Minkowski spaces

A vector field $X$ on a manifold $M$(Riemannian manifold or Lorenztien manifold) corresponds to a process of deformation of this manifold. As in fluid mechanics, there is a corresponding deformation tensor $D_iX_j+D_jX_i$. If the connection on this manifold is Ricci-Civita, then we can show that this tensor is just $L_Xg$, the Lie derivative of the metric tensor with respect to the vector field $X$: $(D_iX_j+D_jX_i)e^ie^j=L_Xg$. And we call $X$ a Killing field if this tensor is identically zero, $D_iX_j+D_jX_i=0$. In other words, the deformation caused by a Killing vector field has no rotation, no dilatations, only some translation(the interpretation in fluid mechanics), or simply is an isometry. We can, for some nice spaces, even calculate the dimension of the vector space spanned by all the Killing vector fields on this manifold. A general fact is that, this dimension is no greater than $n(n+1)/2$ where $n$ is the dimension of the space.

In general, Killing fields are very hard to find, so sometimes we inquire that the deformation be only a conformal differemorphism, which means that the pull-back of the metric is the metric itself multiplied by a scalar positive function. Suppose $\phi_t$ is the local group of deformation determined by the vector field $X$, then we say that $X$ is a conformal Killing field if $\phi^*_t(g)=F(t)^2g$ where $F:M\times (-\epsilon,\epsilon)\rightarrow \mathbb{R}$ is a smooth positive function. Taking derivative with respect to $t$ on both sides at $t=0$, we get that $L_Xg=F(0)F'(0)g$. This means that if $X$ is only a conformal Killing field, then the deformation is a dilatation.The reverse is also true,that is if $L_Xg=ag(a\in\mathbb{R})$, then $\frac{d}{dt}\phi^*_tg=\phi_t^*(L_Xg)=\phi^*_t(a)\phi^*_tg$, then we can solve this differential equation, $\phi_t^*g=e^{\int_0^t\phi^*_s(a)ds}\phi^*_0g=e^{\int^t_0\phi^*_s(a)ds}g$ which is obvious a conformal differemorphism.

In this post, we will work out all the conformal Killing fields on $\mathbb{R}^{1+n}$, the Minkowski space with the metric $m(m_{00}=-1)$.

Suppose that $X$ is a conformal Killing field, that is, $D_{\alpha}X_{\beta}+D_{\beta}X_{\alpha}=L_Xm_{\alpha\beta}=Fm_{\alpha\beta}$.

So we get that $D_{\gamma}D_{\alpha}X_{\beta}+D_{\gamma}D_{\beta}X_{\alpha}=D_{\gamma}Fm_{\alpha\beta}$. After permutations of the indice, $(\gamma,\alpha,\beta)\rightarrow(\alpha,\beta,\gamma)\rightarrow(\beta,\gamma,\alpha)$, add the first two and minus the third one, using the very important, essential fact that in $\mathbb{R}^{1+n}$, $D_{\alpha}D_{\beta}=D_{\beta}D_{\alpha}$, the result is

$D_{\gamma}D_{\alpha}X_{\beta}=1/2(D_{\gamma}Fm_{\alpha\beta}+D_{\alpha}Fm_{\beta\gamma}-D_{\beta}Fm_{\gamma\alpha})$. Contracting the indice $\alpha,\beta$, we get that $\Box X_{\gamma}=1/2(1+1-(n+1))D_{\gamma}F=-(n-1)/2D_{\gamma}F$. Contract the indice in $D_{\alpha}X_{\beta}+D_{\beta}X_{\alpha}=Fm_{\alpha\beta}$, we get that $D^{\alpha}X_{\alpha}=(n+1)/2F$.

Then using the last two formule, we get that $\Box F=0$. From this, we see that $-(n-1)/2D_{\alpha}D_{\beta}F=\Box D_{\beta}X_{\alpha}=1/2\Box (D_{\alpha}X_{\beta}+D_{\beta}X_{\alpha})=1/2\Box Fm_{\alpha\beta}=0$.

In other words, the second derivatives of $F$ is identically zero, which means that $F$ is an affine function on $x^{\alpha}$, $F=a+b_{\alpha}x^{\alpha}$. But $D_{\alpha}X_{\beta}+D_{\beta}X_{\alpha}=Fm_{\alpha\beta}$, which means that each $X_{\alpha}$ is at most a quadratic function on $x^{\beta}$.

Intuitively, what are the possible conformal transformations of $\mathbb{R}^{1+n}$? The translations, the orthogonal transformations, the dilatations, and the inversions. Note that an inversion itself is not a group of conformal transformations parametrized by a real variable, say $t\in(-\epsilon,+\epsilon)$, yet it can act as conjugation on the other one-parameter conformal groups. We can show that, for example, the Killing field corresponding to the translations is $T=a^{\alpha}\frac{\partial}{\partial x^{\alpha}}$, the Killing field corresponding to the orthogonal rotations is $R=x^{\alpha}\frac{\partial}{\partial x^{\beta}}-x^{\beta}\frac{\partial}{\partial x^{\alpha}}$, the Killing field corresponding to the dilatation is $D=x^{\alpha}\frac{\partial}{\partial x^{\alpha}}$, and the Killing field corresponding to the translation conjugated by the inversion is $I_{\alpha}=2x_{\alpha}x^{\beta}\frac{\partial}{\partial x^{\beta}}-x^{\beta}x_{\beta}\frac{\partial}{\partial x^{\alpha}}$.

We can show that these are all the conformal Killing field on $\mathbb{R}^{1+n}$. Indeed, from $F$, we consider $Y=X-aD-b^{\alpha}I_{\alpha}$. After some complicated calculations, we get that $D_{\alpha}X'_{\beta}+D_{\beta}X'_{\alpha}=0$. So $X'$ is itself a Killing field. Yet, we know that the dimension of the vector space of the Killing fields is no more than $(n+1)(n+2)/2$ for $\mathbb{R}^{1+n}$. Yet there are already $n+1$ translations and $(n+1)(n+1-1)/2$ orthogonal rotations, which sum up to $n+1+(n+1)n/2=(n+1)(n+2)/2$! So, this means that $X'$ is a combination of translations and orthogonal rotations, thus $X$ itself is a combination of these four conformal Killing fields. Let’s count the dimension of the vector space of conformal Killing fields, it is $(n+1)(n+2)/2+1+(n+1)=(n+2)(n+3)/2$.