What are the major and immediate consequences of relaxing the condition of positive definiteness axiom of the definition of the inner product? By relaxing, I mean that $\langle x,x\rangle$ is allowed to be any real number, with $\langle x,x\rangle=0$ if and only if $x=0$.

EDIT: Clearly, allowing the inner product to be any complex number would violate the conjugate property <x,x>* = <x,x>,and for this to hold, <x,x> must belong to R. Also,the lack of ordering of the Complex field would mean that we won't be able to compare lengths of vectors.

So, let me correct my question and ask that what happens when we just allow the inner product to be any real number?

Thanks to @gammatester and @Miguel Garcia for pointing these out.

  • $\begingroup$ You cannot always define a unit vector corresponding to $x\ne 0.$ $\endgroup$ Commented Aug 26, 2016 at 10:26

3 Answers 3


I would like to point out that usually the statement "relaxing the condition of positive definiteness" means that there might be non-trivial vectors (i.e. vectors that are not $0$) such that $<x,x>=0$. However it is still true that $<x,x>\in\mathbb{R}$. I do not think it makes much sense to define a dot product that allows $<x,x>\in\mathbb{C}$ since $\mathbb{C}$ is not an ordered field (it does not make sense to say $3+2i>1+i$ and the reason to define a dot product is precisely this, being able to compare lengths of vectors.

On the other hand dot products that are not positive definite are not that strange, in particular Einstein theory of relativity is based on the existence of such a dot product (more precisely the metric tensor). The velocity of light in spacetime is always given by a four vector $x$ satisfying $<x,x>=0$

Hope this helps!

  • $\begingroup$ Thanks! I see your point and have edited the question accordingly. $\endgroup$
    – Razor
    Commented Aug 26, 2016 at 11:03

$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Brak}[1]{\left\langle #1\right\rangle}$If $V$ is a finite-dimensional real vector space, then a symmeric, bilinear, real-valued pairing $\Brak{\ ,\ }$ is non-degenerate if

For every $x$ in $V$, there exists a $y$ in $V$ such that $\Brak{x, y} \neq 0$.

I assume you're asking about non-degenerate pairings.

If the pairing is neither positive-definite $(\Brak{x, x} > 0$ for all $x \neq 0$) nor negative-definite $(\Brak{x, x} < 0$ for all $x \neq 0$) , the quadratic function $Q(x) = \Brak{x, x}$ is continuous and changes sign, hence vanishes for some non-zero $x$.

There are immediate algebraic consequences, such as

A non-zero vector may be orthogonal to itself.

A non-zero vector can fail to be proportional to a unit vector. (See gammatester's comment.)

If we define the unit sphere to be the set of $x$ such that $\Brak{x, x} = 1$, the primary technical consequence of indefiniteness is arguably

The unit sphere is non-compact.

Generally, if $(M, \Brak{\ ,\ })$ is a compact manifold equipped with a continuous/smooth field of non-degenerate, indefinite inner products, then the unit sphere bundle is non-compact. This completely changes the character of, say, the geodesic equations on a compact manifold. Compare, for example:

  • Riemannian geodesics (critical points for the energy with respect to a Riemannian metric, a.k.a., a field of positive-definite inner products), whose long-time existence is guaranteed.

  • Planetary orbits (critical points for the Lagrangian of Newtonian mechanics, which acts as an indefinite metric on the tangent bundle of the configuration space), which can end in collision in finite time.

  • Timelike geodesics in general relativity, which after finite proper time can cease to be extendible.

  • $\begingroup$ Why is the unit sphere not compact in this case? $\endgroup$
    – user297008
    Commented Aug 26, 2016 at 20:00
  • $\begingroup$ @Tanuj: The set of unit vectors is a hyperboloid, disconnected if $\Brak{\ ,\ }$ has signature $(1, n - 1)$ or $(n - 1, 1)$, and connected otherwise. $\endgroup$ Commented Aug 26, 2016 at 21:41
  • $\begingroup$ @Tanuj: Small correction of earlier brain lapse: The set of unit vectors is disconnected if and only if $\Brak{\ ,\ }$ has signature $(1, n - 1)$, i.e., there is a one-dimensional subspace on which $\Brak{\ ,\ }$ is positive-definite, and an $(n - 1)$-dimensional subspace where it is negative-definite. (In the case of signature $(n - 1, 1)$, the "sphere" $Q(x) = -1$ has two components.) $\endgroup$ Commented Aug 27, 2016 at 10:40

One of the immediate consequences would be that the conjugate symmetry axiom of the inner product would be false as if $\langle x,x \rangle$ is allowed to be complex, then $\langle x,x \rangle = (\langle x,x \rangle)^*$ won't be true $\forall x$.

  • $\begingroup$ I cannot see this because $0^{*} = 0$ $\endgroup$ Commented Aug 26, 2016 at 10:39
  • $\begingroup$ he is assuming that the inner product $\langle x, x \rangle$ can be a complex number, so conjugate symmetry would not be hold FOR ALL x, for the case where x = 0 of course it will hold, but the axiom states that the conjugate symmetry must hold for all values of x. $\endgroup$ Commented Aug 26, 2016 at 10:44
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    $\begingroup$ No, only the positive definiteness axiom is relaxed, not conjugate symmetry. Can you show how this may fail? $\endgroup$ Commented Aug 26, 2016 at 10:50
  • $\begingroup$ I know, what I mean is that if you relax that axiom as the question states, then the conjugate symmetry is not hold. I am not relaxing it, it is not hold due to the fact that the $<x,x>\in\mathbb{C}$, implies that there is no conjugate symmetry. It is a consequence, not something that I am assuming. $\endgroup$ Commented Aug 26, 2016 at 10:53
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    $\begingroup$ IMO it says $<x,y>=<y,x>^{*}$ which implies $<x,x> \in \mathbb{R}$ and the positive definiteness axiom is independent. Another question would be: What are the consequences if $\Im(<x,x>) \ne 0$ is allowed. $\endgroup$ Commented Aug 26, 2016 at 11:01

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