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Inner product space is a vector space $V$ over a field $F$ together with an inner product $P:V\times V\to F$ that satisfies the inner product axioms. Let $V_{-0} = V \setminus \{0\}$.

This inner product induces an angle $\angle (v,w) = \arccos \left(\frac {P(v,w)}{||v||\cdot ||w||}\right)$, where $\angle : (V_{-0})\times (V_{-0}) \to [0,\pi]$.

I am wondering if there is an axiomatization of an "angled vector space" $(V,\angle)$ with $\angle:(V_{-0})\times (V_{-0})\to [0,\pi]$, such that

"angled vector space axioms". What are a set of axioms on a pair $(V,\angle)$ such that they are satisfied if and only if there exists an inner product space that induces $\angle$?

Alex M.
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user56834
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  • Note: this question asks a similar question but expressly asks for axioms that are weaker than the inner product axioms, whereas I'm asking for equivalent axioms: $(V,\angle)$: https://math.stackexchange.com/questions/768515/axiomatization-of-angle-measuring-in-real-vector-spaces – user56834 Apr 06 '19 at 16:31
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    It is equivalent to axiomatize the function $\angle'(v,w) = \frac{P(v,w)}{||v||\cdot ||w||}$, where $\angle' : (V/v_0)\times (V/v_0) \to [-1,1]$. – Paul Frost Apr 07 '19 at 17:07
  • @PaulFrost, what do you mean it's "equivalent"? that you'll get the same axioms?? that would surprise me. – user56834 Apr 07 '19 at 17:40
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    Of course not the same axioms, but another set of axioms which can be transformed to axioms (via cos resp. arccos) for an angle function. – Paul Frost Apr 07 '19 at 20:19
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    @user56834: I believe that you should restrict the question to $F = \mathbb R$; otherwise (if $F = \mathbb C$, for instance), the scalar product of two vectors may be non-real, and it is not clear whether its arc-cosine means anything geometric anymore. – Alex M. May 02 '19 at 10:36

2 Answers2

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There may be more "natural" axiomatizations, but you can get an axiomatization that works rather straightforwardly by just reconstructing an inner product (up to scaling) from its angle function (or even just from knowing which vectors are orthogonal).

Indeed, suppose $\angle$ is the angle function of an inner product. We say $v,w\in V$ are orthogonal if either $\angle(v,w)=\pi/2$ or either $v$ or $w$ is $0$. Then note that $\|v\|=\|w\|$ iff $v+w$ and $v-w$ are orthogonal. So, we can recover when two vectors have the same norm, which means we can recover the norm itself up to scaling. We can then recover the inner product by polarization.

So here, then, are some axioms you can use for $\angle$ which are equivalent to it coming from an inner product. For $v,w\in V$, say that $v\sim w$ if $v+w$ and $v-w$ are orthogonal.

  • Axiom 1: $\sim$ is an equivalence relation, and for any nonzero $u\in V$ and any $v\in W$, there is a unique $c\geq 0$ such that $cu\sim v$.

Assuming Axiom 1, for any nonzero vector $u$, we write $\|v\|_u$ for the unique $c\geq 0$ such that $cu\sim v$. We also write $P_u(v,w)=\frac{1}{4}(\|v+w\|_u^2-\|v-w\|_u^2).$

  • Axiom 2: For any nonzero $u\in V$, $P_u$ is an inner product on $V$, and $\angle (v,w) = \arccos \left(\frac {P_u(v,w)}{P_u(v,v)^{1/2} P_u(w,w)^{1/2}}\right)$ for all nonzero $v,w\in V$.

Obviously these two axioms imply that $\angle$ comes from an inner product (namely $P_u$ for any nonzero $u$; I leave the case $V=0$ to the reader). Conversely, if $\angle$ comes from an inner product $Q$, then Axiom 1 is satisfied since $v\sim w$ means $\|v\|_Q=\|w\|_Q$, and Axiom 2 is satisfied since $P_u$ is just $Q$ rescaled so that $u$ is a unit vector.

Eric Wofsey
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  • I agree that this is not "natural" and I was implicitly looking for a "natural" axiomatization. in my opinion, axiom 2 is "cheating" since it uses the axiomatization of inner products basically. axiom 1 is not really cheating but still kind of unnatural, since you're defining a new structure $~$ and use an axiomatization on relations (rather than angles) to state the axiom. – user56834 May 02 '19 at 08:46
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I think that the following should work, but I did not have the time to check all the details.

We use the following, naturally occuring axioms:

  1. Nontriviality: $\angle(v,w) = 0$ iff $v = \lambda \, w$ for some $\lambda > 0$
  2. Summing of angles: $\angle(v,w) + \angle(-v,w) = \pi$
  3. Symmetry: $\angle(v,w) = \angle(w,v)$
  4. Triangle-Equality (triangle with vertices $0$, $v$, $w$): $\angle(v,w) + \angle(-v,w-v) + \angle(-w,v-w) = \pi$
  5. Scaling: $\angle(v,w) = \angle(\alpha \,v, \beta\,w)$ for all $\alpha, \beta > 0$
  6. Iterated law of sines: $$\frac{\sin(\angle(-v,w-v))}{\sin(\angle(-w,v-w))} = \frac{\sin(\angle(-u,w-u))}{\sin(\angle(-w,u-w))} \cdot \frac{\sin(\angle(-v,u-v))}{\sin(\angle(-u,v-u))}$$ whenever all denominators are nonzero.

Now, we define a norm on $V$ by the following procedure: fix $v \in V_0$ and define $\|\lambda \, v\| = \lambda$ for all $\lambda \in \mathbb R$. For an arbitrary vector $w$ which is not in the span of $v$, consider the triangle with vertices $0$, $v$, $w$. From this triangle, we know the length of one side and all three angles. Hence, the triangle can be constructed in the Euclidean plane and we can measure the length of $w$. This leads to $$ \|w\| = \frac{\sin(\angle(-v,w-v))}{\sin(\angle(-w,v-w))} \ne 0.$$

Now, we have to check the axioms of a norm.

  1. $\|w\| = 0$ implies $w = 0$ per construction.
  2. $\|\alpha \, w\| = \alpha \, \|w\|$ follows from a two-dimensional consideration.
  3. Triangle inequality follows from triangle inequality in $\mathbb R^2$.

Finally, the parallelogram identity should follow from a three-dimensional consideration.

gerw
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  • This can't be right: your axioms only depend on the restriction of $\angle$ to each $2$-dimensional subspace. So $\angle$ could be given by different inner products on each $2$-dimensional subspace, which do not combine to give an inner product on the whole space (note that the bilinearity axiom for inner products is a condition on all $3$-dimensional subspaces). – Eric Wofsey May 02 '19 at 14:47
  • Note that in particular the relation $\sim$ as in my answer depends only on the restriction to each $2$-dimensional subspace, so if you pick separate inner products on each $2$-dimensional subspace it can fail to be transitive. That is, given linearly independent $u,v,w$, you could pick an angle function on the plane of $u$ and $v$ which makes them have the same norm, an angle function on the plane of $v$ and $w$ which makes them have the same norm, but an angle function on the plane of $u$ and $w$ which makes them have different norms. – Eric Wofsey May 02 '19 at 14:48
  • Also, one more major thing that you need to check which you haven't mentioned is that $\angle$ actually is the angle function induced by your norm. This plausibly is true essentially by definition for angles in a plane which contains $v$, but it seems you are going to need some $3$-dimensional axiom to verify it in planes that do not contain $v$. – Eric Wofsey May 02 '19 at 15:00
  • I see your point. Maybe we need something like: Consider four points $(0,u,v,w)$. Then, all 12 angles between these point can be realized by a tetrahedron in $\mathbb R^3$. – gerw May 02 '19 at 18:01
  • I have added a "iterated law of sines" (derived by expanding $\frac{|w|}{|v|} = \frac{|w|}{|u|} \cdot \frac{|u|}{|v|}$). I think this should help to reduce every question to a geometric problem in $\mathbb R^3$. – gerw May 03 '19 at 07:10