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I've been reading Fleisch's "A Student's Guide to Vectors and Tensors" as a self-study, and watched this helpful video, also by Fleisch. Suddenly co-vectors and one-forms make more sense than they did when I tried to learn the from Schutz's GR book many years ago.

Especially in the video, Fleisch describes one way of viewing a second-rank tensor; as an object whose individual components relate the input and output components of two sets basis vectors.

In Chapter 4 of his book, Fleisch draws careful attention to two different ways that second rank tensors can be used to relate vectors; they can be used to change a vector, or to change the coordinate representation of a vector.

So, in a classical rigid-body problem, a second-rank tensor is used to create a new vector. The moment-of-inertia tensor operates on an angular velocity vector expressed in terms of a set of basis vectors, and returns an angular momentum vector, which is a different vector that is expressed in terms of the input basis-vectors.

In a change of co-ordinates problem, on the other hand, a second rank tensor operates on a vector expressed in terms of a set of basis vectors, and returns the same vector expressed in terms of a different set of basis vectors.

Even though different things are happening in the two problem types, it is obvious in both cases that the fundamental definition of the problem requires two sets of basis vectors, one used to describe the input and one used to describe the output.


Now, in section 5.5 of his book, Fleisch describes the metric tensor, and describes its function as "allow[ing] you to define fundamental quantities such as lengths and angles in a consistent manner at different locations".

My question is this: Why does this require two sets of basis vectors? If I just want to measure a distance, not change the vector that describes that distance, and not change coordinates, why do I need to involve the second set of basis vectors that a tensor implies? Or, to put the question more concretely, What does the "second set" of basis vectors represent, for example in the Cartesian metric in 3-dimensions?

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  • $\begingroup$ Inner products/metrics not only allows us to define lengths, but also angles between two vectors. $\endgroup$ – SCFT Jun 16 '17 at 5:12
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Assuming positive definite metrics for simplicity (so no spacetime), your "metric tensor" can actually be considered as a single-variable object. For a vector $v$, it outputs a number $Q(v)$, which intuitively corresponds to the magnitude of a vector. This satisfies the following rules:

  • Positive definiteness: $Q(v)\ge 0$, with $=$ if and only if $v=0$.

  • Absolute-value homogenity: $Q(\alpha v)=|\alpha|Q(v)$ for a real number $\alpha$.

  • Triangle inequality/subadditivity: for vectors $v,w$ $Q(v+w)\le Q(u)+Q(v)$.

What this gives is actually a norm. A geometry consisting of a smooth manifold $M$ and an assignment of such a norm $Q_x$ to every point $x\in M$ (such that this norm varies smoothly) gives what is called Finsler-geometry.

A norm contains less information than a metric tensor though (and is not a tensor in general), because while a norm gives lengths/distances, it cannot describe angles.

A norm may or may not satisfy the so-called parallelogram rule, which states that for every vectors $v,w$ we have $$ 2Q(v)^2+2Q(w)^2=Q(v+w)^2+Q(v-w)^2. $$ It can be shown, that if a norm satisfies this identity, then there exists a unique metric tensor $g$, with the following properties:

  • For any $v$ vector $g(v,v)=Q(v)^2$.

  • For any two vectors $v,w$, we have $g(v,w)=\frac{1}{4}(Q(v+w)^2-Q(v-w)^2)$ (polarization identity).

The metric tensor then has the following geometric interpretation: $g(v,v)=Q(v)^2$, so $g(v,v)$ gives the square of $v$'s length, and $g(v,w)=Q(v)Q(w)\cos\alpha$, where $\alpha$ is the angle between the two vectors.

So to actually answer your question, the metric tensor's length calculation property is actually given by a single-variable object expressible as $Q(v)=\sqrt{g(v,v)}$, but because a metric tensor can also calculate angles, it also has a representation, which take in two vectors, because two vectors are needed to give an angle.

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  • $\begingroup$ Building on the last part of @Uldreth's answer, i.e. "because two vectors are needed to give an angle"... Why does the simplest form for defining angles require that two sets of basis vectors be provided (implicitly, by the specification of the tensor) . IOW, for defining lengths and angles, etc., why doesn't a mathematical form that assumes that one set of basis vectors will be used everywhere suffice? $\endgroup$ – NaiveBayesian Jun 17 '17 at 15:46

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