In learning about (Einstein) tensor index notation I've noticed what I feel are inconsistensies with the notation. Now in Einstein notation vectors are denoted by superscripts $v^i$ while co-vectors are denoted by subscripts $v_i$. However in the context of differential forms, the co-vectors $dx^i$ are denoted with superscripts and the vectors $\frac{d}{dx^i}$ by subscripts which is opposite to Einstein summation convention -- furthermore in the same (differential-geometric) context the geometric tensor is written as $g_{ij}$ which is in accordance with the Einstein convention(on account of it being a $(0,2)$ tensor). However, this seems to make expressions such $ds^2 = g_{ij}dx^idx^j$ very odd, since while this is clearly what one would want in order get the right sum by contracting indices it nonetheless is extremely weird that we seem to be essentially contracting co-vectors by other co-vectors.

In any case, I wonder is the notation used in fact inconsistent when it comes to denoting co-vectors by subscripts/superscripts; and if so, where did this inconsistency originate from?

  • 3
    $\begingroup$ No no, the coefficients of vectors are denoted by superscripts and the coefficients of co-vectors are denoted by subscripts. $\mathrm dx^i$ is a very natural notation, since it is simply the differential of the map $x^i$ on your smooth manifold. Similarly, $\frac{\partial}{\partial x^i}$ denotes the "derivative with respect to the $i$-th coordinate $x^i$". $\endgroup$ Sep 16, 2021 at 11:44
  • 1
    $\begingroup$ Put differently: $g_{i,j}$ is by itself just a coefficient of the $2$-covariant tensor $g$. The latter can be written out as $$g = \sum_{i,j=1}^{n} g_{i,j}\ \mathrm dx^i\otimes\mathrm dx^j,$$ where once again $\mathrm dx^i, \mathrm dx^j$ are the differentials of the functions $x^i$ and $x^j$. (And $n$ is the dimension of your manifold, the expression only works on the domain of your chosen chart $x=(x^1,\dots, x^n)$, cf. the chapter 1 of Lee's Introduction to Smooth Manifolds, 2012 for the definition of chart that I am using here.) $\endgroup$ Sep 16, 2021 at 11:48

1 Answer 1


You need to differentiate between the components of a vector and the vector itself.

The $n$ coordinate vectors $\partial_j$ constitute a basis of the n-dimensional tangent space $T_p(M)$ at each point $p\in M$. Sometimes these coordinate bases induced by the charts $(U,h)$ are referred to as the natural or canonical bases.

Relative to a given chart (U,h) with coordinate functions $x^1,\dots, x^n$ any vector $v\in T_p(M)$ admits the representation $$v=v^i\frac{\partial }{\partial x^i}$$

When we say that $v^i$ is a contravariant vector we actually mean that the components $v^i$ transforms like a contravariant vector at $p$ (and that $v\in T_p(M)$). The coefficients are uniquely defined by $$v^i=vx^i$$

For two overlapping charts the vector $v$ is represented by $v=v^i\frac{\partial}{\partial x^i}$ and $v=\bar{v}^j\frac{\partial}{\partial {\bar{x}}^j}$ respectively. With $v^i=v x^i$ and $\bar{v}^j=v\bar{x}^j$ we can make two important observations $$\bar{v}^j=v^{i}\frac{\partial \bar{x}^j}{\partial x^i}\tag{1}$$

$$v=\bar{v}^j\frac{\partial }{\partial \bar{x}^j}=v^h\frac{\partial\bar{x}^j}{\partial x^h}\frac{\partial}{\partial \bar{x}^j}=v^h\frac{\partial}{\partial x^h}\tag{2}$$

Now $(1)$ tells us that indeed the components $v^i$ transforms like the components of a contravariant vector. Further, since $(2)$ is valid for any arbitrary $v^i$ we conclude that

$$\frac{\partial }{\partial x^h}=\frac{\partial \bar{x}^j}{\partial x^h}\frac{\partial }{\partial \bar{x}^j}$$

So the basis elements for the (contravariant) tangent space itself transform like covariant vectors!

I will not go into details on how to construct the dual basis $\{dx^k,\,k=1\dots n\}$ of $T^*_p(M)$ but it follows naturally from the definition of the unique element $df\in T^*_p(M)$ such that $\langle d f,v\rangle=vf$ that any $\omega\in T^*_p(M)$ can be expressed as


Notice again that $\omega_j$ are the components of a covariant tensor, while $dx^j$ are the basis (which transforms contravariantly). The dual tangent space $T^*_p(M)$ is called the cotangent space, and its elements $\omega$ are referred to as covectors or 1-forms.

Specifically with $df=f_hdx^h$ we have $df=\partial_jfdx^j$ (since $\langle df,\partial_j\rangle=f_j$) which is consistent with the customary expression for the differential of a function.

  • $\begingroup$ This didn't help much either :( $\endgroup$ May 26, 2022 at 21:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .