in $ds^2 = g_{\mu \nu} dx^\mu dx^\nu$, why is $dx^\mu$ represented as a row vector and $dx^\nu$ represented as a column vector?

Question

Just like what's in the title, in the equation $ds^2 = g_{\mu\nu} dx^\mu dx^\nu$, why are $dx^\mu$ and $dx^\nu$ represented differently?

Answer 1

I wouldn't say they are represented by a row and column vectors, instead, they are something more fundamental than how we represent them. You will also need to give more information on the form of $g_{\mu\nu}$ for a more detailed answer. If we use the $(+1,-1,-1,-1)$ metric from special relativity for simplicity of explanation, then $g_{\mu\nu}dx^\mu$ is some covariant vector that we can define as $V_\nu$. Then $g_{\mu\nu}dx^\mu dx^\nu=V_\nu dx^\nu$ which is the sum over component-wise multiplication, we can represent this as: $$\begin{pmatrix} dx^0 &-dx^1 &-dx^2 &-dx^3 \end{pmatrix} \begin{pmatrix} dx^0\\ dx^1\\ dx^2\\ dx^3\\ \end{pmatrix}$$

The result is the same but this loses meaning very quickly for higher-order tensors and abstract basis vectors. Further, if we were looking at the scalar: $g_{\mu\nu}dx^\mu dy^\nu$, then if we defined $V_\nu=g_{\mu\nu}dx^\mu$ or $W_\mu=g_{\mu\nu}dy^\nu$ then the two row/column matrices we multiply switch orders but the final result is the same.

Edit: Further, you may often see lower-dimensional tensor contractions as matrices of the form: $$\begin{pmatrix}x_1&x_2\end{pmatrix}\begin{pmatrix}g_{11}&g_{12}\\ g_{21}&g_{22}\end{pmatrix}\begin{pmatrix}y_1\\ y_2\end{pmatrix}=\begin{pmatrix}x_1&x_2\end{pmatrix}\begin{pmatrix}g_{11}y_1+g_{12}y_2\\ g_{21}y_1+g_{22}y_2\end{pmatrix}$$ $$=\begin{pmatrix}g_{11}x_1+g_{21}x_2&g_{12}x_1+g_{22}x_2\end{pmatrix}\begin{pmatrix}y_1\\ y_2\end{pmatrix}$$ The last two lines are due to the associativity of matrix multiplication and emphasizes the point I made above by contrasting $V_\nu$ and $W_\mu$.

Answer 2

This is called "Eintein's notation". For quick reference, take a look at the article https://en.wikipedia.org/wiki/Einstein_notation. A brief explanation is:

A column vector is given by the summation $\sum\limits_{i=1}^n x_i e_i$, for canonical vectors $e_i$ also represented by delta vector notation $\delta_i$. Einstein noticed we may use upper and lower indices to avoid the summation and spare some tedious repetitive work. It means, column vector ENTRIES are row elements on the column, which takes the upper index notation. The row vector ENTRIES are column elements, which assign them lower indices. Therefore, a vector $v$ is given by notation $v^i e_ i$.

The inner product on $\mathbb{R}^n$ of vectors $v$ and $w$ is given by vectors $v^i w_ i$ because the terms are all summed up and the indexes "collapse".

Here comes your example, which assumes a metric space with metric tensor $g_ij$. Tensors are denoted but double lower indexes $T_{ij}$ and its inverse $T^{ij}$. A 2-form $T_{ij} \, x^i \, \otimes \, x^j$, or in a simpler space, $T_{ij} \, x^i \, x^j$ is somewhat a form of space, a subspace if you like. The notation here is to cmoply with tensor notation, and corresponds to product form, $x^\intercal \, G \, x$.