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If one considers a smooth manifold, then it is possible to define (tangent) vectors in the tangent space to each point on that manifold without having to introduce any additional structure such as a metric.

As such (tangent) vectors can be defined in terms of equivalence classes of curves (with a tangent vector at a given point defined as an equivalence class of curves through that point that are all tangent to one another at that point), is it correct to say the notion of a vector having "direction" is meaningful without needing to introduce a metric, or do the two come hand in hand?

I ask as I know that vectors (in general) do not have "length" without first defining a metric, however, I'm not sure that the same can be said for direction? I think I'm slightly confused with the notion of defining vectors on a manifold and more abstract vector spaces, for example, the set of polynomials form a vector space however it doesn't seem to make sense (at least to me) that these polynomials have "direction"?!

Any help would be much appreciated.

Will
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  • Once you have an inner product, you get length, distance, angle between vectors (relative direction), etc. A vector space on its own isn't endowed with a concept of direction. You have to abstract that graphically somehow (e.g. by identifying polynomials with their coefficient vectors - but at that point we have gone beyond what is required to be a vector space). – jdods Oct 21 '15 at 11:49
  • @jdods That's what I thought. So is it correct to say that in the general case (i.e. abstract vector space) a vector does not have direction or magnitude until one introduces the structure of an inner product? – Will Oct 21 '15 at 12:19
  • I would say that a vector does not have a "direction" until such a concept is defined. We could probably define "direction" without using an inner product -- though I have not done so. For example for $R^2$, let the direction of a vector be the difference between its coordinates and the direction between two vectors be the difference between their directions: dir(u)=|u_1-u_2|, dir(u,v)=dir(u)-dir(v). Intuitively, I don't think this would satisfy all the requirements of inner product, distance, measure, etc. – jdods Oct 21 '15 at 21:41

3 Answers3

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Given a real vector space $V$, e.g. the tangent space at the point $p$ of a $d$-dimensional manifold $M$, call two nonzero vectors $a$, $b\in V$ equivalent if $b=\lambda a$ for some $\lambda>0$. An equivalence class is then a direction in $V$. A scalar product $\cdot$ on $V$ defines a norm, and in the sequel a metric, on $V$ via $|a|^2:=a\cdot a$, but also a metric on the set $S^{d-1}$ of directions via $$d\bigl([a],[b]\bigr):=\arccos{a\cdot b\over|a|\>|b|}\ .$$

Christian Blatter
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  • So in terms of vectors on a manifold is it their definition as equivalence classes of curves on the manifold that gives a notion of them having "direction"? – Will Oct 21 '15 at 12:17
  • Any nonzero vector "has" a direction, meaning: belongs to an equivalence class in the above sense. – Christian Blatter Oct 21 '15 at 13:18
  • I'm confused really in the case of introducing vectors in differential geometry. These can be defined as equivalence classes of curves that are mutually tangent at a particular point. Is this what you mean by a vector having direction in terms of the equivalence class you specify in your answer? – Will Oct 22 '15 at 10:36
  • No; see my edit. – Christian Blatter Oct 22 '15 at 10:51
  • Ah ok, so is the point that the equivalence class of curves defines what a vector is, but does not specify that it has any notion of direction, and then the equivalence class of vectors (defined as above in your answer) is an equivalence class in the tangent space, between vectors in that tangent space, that defines an elementary notion of a vector having direction. Is that correct at all? – Will Oct 22 '15 at 11:03
  • Note that this defines what might be thought of as "relative direction". A vector can also be thought of having an intrinsic sense of direction. – jdods Oct 22 '15 at 12:36
  • @jdods In what sense can they have an intrinsic sense of direction? One can construct an abstract vector space without mention of the vectors having direction (other than in the relative sense described in the answer above). – Will Oct 22 '15 at 13:40
  • See my answer below. It's not based on rigorous ideas, but a vector can be thought of as defining a direction itself. That's a natural way to think about it potentially. – jdods Oct 22 '15 at 22:45
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As I noted in my comment, we are probably free to define "direction" however we want as a vector space only satisfies a specific set of axioms that usually don't mention the word "direction". And how we define direction is completely up to us. But...

According to Wolfram Mathworld, direction is a vector itself. This makes intuitive sense to me. And normally it is considered as a unit vector, but you need a concept of distance to have unit vectors.

So if $\mathbf{u}$,$\mathbf{v}\in V$ a vector space, then $\mathbf{v}-\mathbf{u}$ is the direction from $\mathbf{u}$ to $\mathbf{v}$. This 'direction vector' automatically defines an equivalence class of vectors: $\mathbf{a}+\mathbf{v}-\mathbf{u}$ is the same direction from $\mathbf{a}$ as $\mathbf{v}$ is from $\mathbf{u}$. Of course, this is different from the concept of angle since it disregards distance.

I'd be curious to see what advanced texts say about the concept of direction before introducing inner products.

jdods
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What about the following definition of "direction" that one can use without appealing to any additional structure besides the vector space structure? (I.e. no need for inner products.)

(Note that, compared to jdods answer, this only gives "direction" relative to the origin/additive identity of a vector space, not relative to an arbitrary vector. It would be possible to extend this definition so that it gave direction relative to an arbitrary vector in $V$ -- just use the tangent space.)

Let $V$ be a vector space over the field $F$ with dimension greater than or equal to $1$.

  • By convention, $0 \in V$ has no direction.
  • If $F$ is an ordered field, then two non-zero vectors $v_1, v_2 \in V$ have the same direction if and only if $$ v_1 = \lambda v_2 \quad \text{for some }\lambda>0 \iff v_2 = \frac{1}{\lambda}v_1 \quad \text{for some }\frac{1}{\lambda}>0\,. $$
  • If $F$ is not an ordered field, then two non-zero vectors $v_1, v_2 \in V$ have the same direction if and only if $$v_1 = \lambda v_2 \quad \text{for some }\lambda \not=0 \quad v_2 = \frac{1}{\lambda}v_1 \quad \text{for some }\frac{1}{\lambda}\not=0\,.$$

Note that, for a finite dimensional vector space $V$ over a non-ordered field $F$, i.e. $V \cong F^n$ for some $n\in \mathbb{N}, n \ge 1$, the definition is equivalent to the two vectors belonging to the same equivalence class of the equivalence relation which defines $\mathbb{P}^{n-1}(F)$, the projective space of dimension $n-1$ over $F$.

The definition in the case of an ordered field does not correspond neatly to the definition of projective space, even though the definition is admittedly similar, it's not the same.

Note also that this definition is the same one given in Christian Blatter's answer when $F = \mathbb{R}$. But one doesn't need the underlying field to be $\mathbb{R}$ in order to make a sensible definition work.

Chill2Macht
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