How to find the standard Euclidean metric in $\mathbb{R}^2$ given the metric tensor?

Question

I have seen in texts that the line element $ds$ can be shown to be $$ds^2=g_{ij}\,dx^i\,dx^j$$ where summation is implied. I tried to use this formula to verify the Eulcidean metric on $\mathbb{R}^2$ on the standard $x$-$y$ Cartesian plane. I know that the metric tensor in this situation is given by $$g=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$ To actually simplify the right-hand side, does the notation mean that I multiply each component of the matrix by the respective differential? In this case, it would be $$ds^2=1\cdot dx^2+0\cdot dx\,dy+0\cdot dy\,dx+1\cdot dy^2=dx^2+dy^2$$ I see that this is correct, but I am wondering if there is another method that this process can be carried out by. For example, can we represent this as the multiplication of two matrices?

Answer 1

Yes, this is usually interpreted as Einstein summation notation, and your expansion is correct. If you like, you can rearrange it to $$ds^2 = \sum_{i,j}dx^i\, g_{ij} \,dx^j,$$ which is the component formula for the formal matrix product $dx^T \,g \,dx$ where $dx$ is the column vector with entries $dx^i$. (Depending on how you're interpreting $dx^i$ this may or may not be an abuse of notation.) If you know some linear algebra this should be familiar as the action of the quadratic form $g$ on the "vector" $dx$.