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Since vectors in general relativity are represented as directional derivative operators, it seems to me that the product of two vectors should be a second order derivative operator. However, the product of two vectors is not an operator, but a scalar number just like a scalar product in linear algebra. Why?

Qmechanic
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HelloGoodbye
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2 Answers2

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Vectors are represented as differential operators like $$ v = v^\mu \partial_\mu. $$ These are spanned by a basis of $\partial_\mu$.

In addition to vectors, there is also something called a "1-form."

$$ \omega = \omega_\mu dx^\mu. $$ These are spanned by a basis of $d x^\mu$.

In particular, these 1-forms can be thought of as a map

$$ \omega : \text{tangent vectors} \to \text{scalars} $$

For instance, $$ \omega (v) = \omega_\mu v^\mu. $$ In particular, $$ d x^\mu ( \partial_\nu) = \delta^\mu_\nu. $$ and $$ dx^\mu (v) = v^\mu. $$

Now, the metric is made up of a sum of symmetric products of these one-forms. (It is not a "2-form" because it is made of symmetric products, not anti-symmetric products.)

$$ g = g_{\mu \nu} dx^\mu \otimes dx^\nu. $$ Usually the symbols $\otimes$ is dropped, but whatever. Anyway, the metric is a map $$ g : \text{two tangent vectors} \to \text{scalars} $$ and acts as \begin{align} g(v, u) &= g_{\mu \nu} dx^\mu \otimes dx^\nu (v, u) \\ &= \frac{1}{2} g_{\mu \nu}(( dx^\mu (v) )( dx^\nu (u) )+( dx^\mu (u) )( dx^\nu (v) )) \\ &= \frac{1}{2} g_{\mu \nu}( v^\mu u^\nu + u^\mu v^\nu) \\ &= g_{\mu \nu} v^\mu u^\nu \end{align}

so that is really just how everything is defined.

user1379857
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  • Thanks, that makes sense. I often see the notation $u\cdot v$. Is this defined as $g(u,v)$, in contrast to simply being the product of $u$ and $v$, i.e., $uv$? – HelloGoodbye Oct 02 '21 at 13:21
  • I have not really seen that notation but I can't imagine what else $u \cdot v$ could be. There aren't many options. – user1379857 Oct 02 '21 at 21:06
  • It is used for example in Carroll's Spacetime and Geometry: An Introduction to General Relativity. It's the same as the notation for the dot product used in linear algebra. – HelloGoodbye Oct 02 '21 at 21:32
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    "The action of the metric on two vectors is so useful that it gets its own name, the inner product (or scalar product, or dot product): $\eta(V,W)=\eta_{\mu\nu}V^\mu W^\nu=V\cdot W.$" – HelloGoodbye Oct 02 '21 at 21:37
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In differential geometry, tangent vectors can also be represented by partial derivatives, but nevertheless they are also vectors as traditionally understood. The reason for alternative description by derivative operators is that we can describe tangent vectors intrinsically. However, we can see that tangent vectors are tangent vectors by simply embedding the manifold in a larger Euclidean space. Hence the scalar product is defined as traditionally done and is a scalar.

Mozibur Ullah
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  • What does it mean to be able to describe a vector intrinsically? And what do you mean by "hence the scalar product is defined as traditionally done"? In Linear algebra, defining a scalar product is straightforward because you can easily pick out the individual elements, but it's not obvious how to do that for a vector in GR unless you also has the component tensor. – HelloGoodbye Oct 02 '21 at 13:27
  • @HelloGoodbye: A tangent vector is described intrinsically when it requires nothing more than the information required to construct the manifold. It's described extrinsically when more information is required, this is generally when the manifold is embedded in an ambient Euclidean space. – Mozibur Ullah Oct 02 '21 at 23:09
  • @HelloGoodbye: When the tangent vector is described extrinsically, we have two descriptions available - the traditional and the differential operator descriptions. This is best seen geometrically: think of a surface in 3d space and draw two tangent vectors at a point. You can easily see we have the scalar product as traditionally described. Now think of a 7d manifold, this can be embedded, in say a 21d Euclidean space, then we can do the analogous thing as before and use the traditional scalar product. – Mozibur Ullah Oct 02 '21 at 23:17