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I am just being introduced to Riemannian metrics, and I am having a bit of confusion on the notation. When reading, I've encountered some different notation in different sources, so I want to make sure I'm understanding what the different notation means. I'm sure there are lots of misunderstandings in what follows, so I'd appreciate if someone could correct them.

Say for example we have $\mathbb{R}^2$ in canonical coordinates, and we're going to use the usual dot product $\langle x,y \rangle = x_1y_1 + x_2y_2$, where $x=(x_1,x_2)$ and $y=(y_1,y_2)$. Then the Riemannian metric would be the $2 \times 2$ identity matrix, so that $$\begin{pmatrix} x_1 & x_2 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} y_1 \\ y_2 \end{pmatrix} = x_1y_1+x_2y_2,$$ so $g_{ij} = \delta_{ij}$. This notation makes sense to me, but I am confused about the notation $g = dx^2+dy^2$. Do $dx$ and $dy$ denote linear functionals in the dual space? And if they do, then is $dx = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $dy = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$? This doesn't seem right, because the dimensions of the matrices wouldn't come out correctly.

Then say we want to convert this to polar coordinates, so $x=r \cos \theta$ and $y=r \sin \theta$. Then do we have $$g = \begin{pmatrix} 1 & 0 \\ 0 & r^2 \end{pmatrix} = dr^2 + r^2 d \theta^2?$$ Then to get the inner product, we would do $$\begin{pmatrix} r_1 & \theta_1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & r^2 \end{pmatrix} \begin{pmatrix} r_2 \\ \theta_2 \end{pmatrix} = r_1r_2 + r^2 \theta_1 \theta_2?$$ If this is correct, then what does the $r^2$ in the metric represent in the calculation? Is it the $r$ that corresponds to the point at which we're defining the metric? Like, if we wanted $\langle v,w \rangle$ where $v = (3,\pi/2)$ and $w = (5,\pi/4)$, what does the expression $$\langle v,w \rangle = 3 \cdot 5 + r^2 \cdot \dfrac{\pi}{2} \cdot \dfrac{\pi}{4}$$ mean, intuitively?

Thanks in advance for the help.

P.S. Other examples, or links to well-explained other examples would be much appreciated.

phaiakia
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1 Answers1

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The $g=dx^2 + dy^2$ notation really means $g = dx \otimes dx + dy \otimes dy$, using the tensor product. Tensor products build up linear maps. That way, you get a map that takes in two vectors to return a scalar--exactly what you would want the metric to do.

In other coordinates (like polar), you must remember to keep a distinction between positions and tangent vectors. The row and column vectors you've written correspond to tangent vectors: elements of a vector space spanned locally by (1) the radial vector and (2) the angular vector. Both of these are evaluated at a point, and the coordinates of that point give you the $r$ that appears in the metric.

Muphrid
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  • I'm using do Carmo's Riemannian Geometry, and he does not discuss tensors before Riemannian metrics. I've only seen tensor products in the context of abstract algebra... could you elaborate a bit on the tensor product notation, or recommend a good source to learn more about it? – phaiakia Jul 20 '14 at 17:54
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    The tensor product obeys the following: $(A \otimes B)(c, d) = A(c) \otimes B(d)$. For the metric, you would have $c,d$ be vectors, and you'd have, for instance, terms like $dx(c) \otimes dx(d)$. But $dx(c)$ is a scalar, and the same for $dx(d)$, so the tensor product reduces to scalar multiplication there. – Muphrid Jul 20 '14 at 17:56
  • Ah, ok. That makes sense. Was I correct in thinking that $dx$ is a linear functional in the dual space? – phaiakia Jul 20 '14 at 18:02
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    Yes, that is precisely what is meant here, that $dx$ is the coordinate basis one-form corresponding to the coordinate $x$. – Muphrid Jul 20 '14 at 18:09