# What's the point of this killing vector notation?

Reading Sean Carroll's spacetime and geometry he says

If $$x^{\sigma_*}$$ is the coordinate which $${\mu\nu}$$ is independent of, let us consider the vector $$\partial_{\sigma_*}$$ which we label as $$K=\partial_{\sigma_*}\, , \tag{3.169}$$ which is equivalent in component notation to $$K^\mu=(\partial_{\sigma_*})^\mu = \delta^\mu_{\sigma_*}. \tag{3.170}$$

Now I ask you, why call a vector that for example looks like this $$(1,0,0,0)$$ as $$\partial_0$$?

Is there a point and a reason behind this notation which I always found pretty confusing? For example before reading Sean Carroll the times I read a phrase like the Killing vector is $$\partial/\partial t$$ I thought it was something like $$(\partial_t,0,0,0)$$ and beside that I was never sure if it was really $$(\partial_t,0,0,0)$$ or if it was a spacelike vector with a component equal to $$\partial_t$$

Just think about ordinary vectors. The vector $$\mathbf{r} = \hat{\mathbf{x}}$$ has components $$(1, 0, 0)$$ in Cartesian coordinates. This is because the components are the numbers you have to multiply the unit vectors with to recover the vector you want, $$\mathbf{r} = 1 \hat{\mathbf{x}} + 0 \hat{\mathbf{y}} + 0 \hat{\mathbf{z}}.$$ The exact same thing is true here, except that the unit vectors are $$\partial_t$$, $$\partial_x$$, $$\partial_y$$, and $$\partial_z$$. Again, the components of vectors are numbers, not vectors themselves.

• Wait, it's because we work in tangent space right? And that's the reason the unit vectors are the partial derivatives? I'm very rusty, sorry if it's a bit trivial. Feb 6, 2020 at 19:12

There are several different ways to define tangent vectors, eg in terms of coordinates and transformation laws, or as equivalence classes of differentiable curves in first-order contact. In differential geometry, we tend to identify tangent vectors with their corresponding directional derivatives, which can be used as a third way to introduce the tangent space.

Writing the basis vectors induced by a coordinate chart as partial derivatives makes sense under that identification.

For example, let $$\Psi = \Psi(x^1,\dots,x^n)$$ be a parametrization of our manifold $$M$$. Then, each of the $$n$$ coordinates will induce a tangent vector $$\partial/\partial x^i$$, corresponding to the directional derivative

$$\frac{\partial}{\partial x^i}f = \frac{d}{d\tau}\Big|_{\tau=0} (f\circ\Psi)(x^1,\dots,x^{i-1},x^i + \tau,x^{i+1},\dots,x^n)$$ of a function $$f:M\to\mathbb R$$.

In terms of this basis, an arbitrary vector $$X$$ can then be expanded as $$X = \sum_{i=1}^n X^i \frac{\partial}{\partial x^i}$$ For $$x^0 = t$$ and $$X=\partial/\partial t$$, we would then have $$X = 1\cdot \frac{\partial}{\partial t} + 0\cdot\frac{\partial}{\partial x^1} + 0\cdot\frac{\partial}{\partial x^2} + 0\cdot\frac{\partial}{\partial x^3}$$ ie $$(X^\mu) = (1,0,0,0)$$ but not $$(X^\mu) = (\partial/\partial t,0,0,0)$$

It comes from the formal definition of a vector on a manifold. Given a manifold $$\mathcal{M}$$ and a chart $$(U,x^{\mu}:U\to\mathbb{R}^n)$$ one defines a curve $$c$$ and a function $$f$$ both differentiable $$c:\mathbb{R}\to\mathcal{M}\,,\qquad f: \mathcal{M}\to\mathbb{R}\,.$$ Since $$f\circ c$$ is a function $$\mathbb{R}\to\mathbb{R}$$ one can define its first derivative $$\left.\frac{\mathrm{d}f(c(t))}{\mathrm{d}t}\right|_{t=0} = \frac{\mathrm{d}x^\mu(c(t))}{\mathrm{d}t} \frac{\partial f}{\partial x^\mu}\equiv V f\,.$$ The differential operator $$V = \frac{\mathrm{d}x^\mu(c(t))}{\mathrm{d}t} \frac{\partial }{\partial x^\mu}$$ is what we call a vector.

The nice thing is that such an expression is coordinate independent: $$Vf$$ only depends on the choice of the curve $$c$$ (modulo an obvious equivalence relation) and $$f$$. On the other hand, when we strip the $$\partial/\partial x^\mu$$ and define $$V^\mu = \frac{\mathrm{d}x^\mu(c(t))}{\mathrm{d}t}\,,$$ we are introducing the dependence on the chosen coordinate frame $$x^\mu$$.