# Covector basis derivation

On page 65 of Schutz's A first course in General Relativity, he introduces the notation $$\phi_{,\alpha}=\partial\phi/\partial x^\alpha$$. He then says that $$x^\alpha_{\ \ ,\beta}=\delta^\alpha _{\ \ \ \beta}$$. I understand this, since it's just another way of writing $$\partial x^\alpha/\partial x^\beta=\delta^\alpha _{\ \ \beta}$$. What I don't get is what he says immediately after: since, given a basis $$\left\{e_i\right\}$$, we can define a dual/covector basis $$\left\{e^j\right\}$$ by $$e^j(e_i)=\delta^j_{\ \ \ i}$$, then $$x^\alpha_{\ \ ,\beta}=\delta^\alpha _{\ \ \ \beta}$$ means that $$\left\{\textrm{d}x^\alpha\right\}$$ is the covector basis. I know it is, but I don't get why it follows from this statement. I would get it, if it were $$\text{d}x^\alpha_{\ \ \ ,\beta}=\delta^\alpha _{\ \ \ \beta}$$ (note the $$\text{d}x$$ rather than simply $$x$$ on the LHS); this would then mean that $$\partial\textrm{d}x^\alpha/\partial x^\beta=\delta^\alpha_{\ \ \ \beta}$$ and so I see why that would be the basis.

Second, $$\{ e_i\}$$ (in your case) represents the set of "coordinate (holonomic)" basis (i.e., $$\{ e_i\}=\{ \partial_i := \partial/\partial x^i \}$$) and not any basis.

You should know a little bit about differential geometry: Given a (smooth) manifold $$M$$ there is a set of basis vector fields $$\{ e_i\}$$ at a point $$x \in M$$ in the tangent space $$T_xM$$, i.e., any vector field $$V \in T_pM$$, at a point $$p$$, can be written as $$V = V^i \left( \frac{\partial}{\partial x^i} \right)_p \,, \quad \text{where} \quad V^i=V(x^i).$$ Now, one can define the dual space of $$T_xM$$, i.e., the cotangent space $$T^\ast_xM := (T_xM)^\ast$$. That is, the elements of $$T^\ast_xM$$ are linear functionals on $$T_xM$$, at each point $$x \in M$$. This means that for $$\omega \in T^\ast_xM$$ we have a linear map $$\omega: T_xM \rightarrow \mathbb{R} \,,$$ since our field in GR is the set of real numbers.

Now, if we denote by $$\{ e_i \}=\{\partial_i\}$$ the set of basis on $$T_xM$$, and its dual by $$\{e^j\} = \{ dx^j \}$$ on $$T^\ast_xM$$ for $$i,j=1,2, \ldots,\text{dim}M$$, there is a canonical isomorphism between these two spaces by the relation $$dx^j (\partial_i) = \delta^j_i \,.$$ So the covector $$dx^j$$ acts as a linear functional on the vector $$\partial_i$$. This relation determines the corresponding dual basis uniquely (provided the space at hand is finite dimensional). Of course, the statements I made above are valid at each point $$x \in M$$, but you can generalise these to tangent bundle, cotangent bundle and so on, and you can simply write the vectors (covectors) as vector (covector) fields, without specifying the point.

For instance, for $$V \in TM$$ (tangent bundle) and $$\omega \in T^\ast_xM$$, we have $$\omega(V) = \omega_i dx^i (V^j \partial_j) = \omega_i V^j \, dx^i (\partial_j) = \omega_i V^j \, \delta^i_j = \omega_i V^i \,.$$ Sorry for compactness of my answer!

• Isn't the isomorphism between the tangent space and the dual space the inner product $\langle dx^j, e_{i}\rangle= \delta^j_i$ instead of $dx^j (x_i) = \delta^j_i\;$? – Cinaed Simson Oct 26 at 6:39
• @Cinaed Simson Of course! It was a taypo, now corrected. – Astrolabe Oct 26 at 7:35

I would get it, if it were $$\text{d}x^\alpha_{\ \ \ \beta}=\delta^\alpha _{\ \ \ \beta}$$ ...

Note that the object $$dx^\alpha$$ is a co-vector, so it cannot have two indices, only one. From the normal rules of differentiation we can write $$dx^\alpha=\frac{\partial x^\alpha}{\partial x^\beta}dx^\beta$$ so the components of the $$dx^a$$ covector in the $$\{dx^\beta\}$$ basis are $$\frac{\partial x^\alpha}{\partial x^\beta}=\delta^\alpha{}_\beta$$, satisfying the definition.