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I read that the reason we have the property $\langle A|B\rangle=\langle B|A\rangle^*$ is to make define a positive norm with the formula $\langle A|A\rangle$.

But I do not understand how having this norm benefits us. I guess we're doing this to make an analogy with arrows, which also have a positive norm.

But this can't be the only reason. After all, a lot of things which are true for arrows are not true for general vectors. For instance, angle values of $-2\pi$ to $2\pi$ are not carried over from arrows to general vectors. The formula for the angle between general vectors, $\cos \theta=\frac{\langle A|B\rangle}{|A||B|}$, can result in complex values of $\theta$. The commutativity of the inner product isn't carried over from arrows to general vectors either (though this is the very reason a positive norm gets carried over).

Keeping a positive norm for general vectors must be allowing us to carry over some nice properties from the world of arrows to general vectors. What are those nice things?

Like, even if we drop this property, we'd still be able to prove the existence of an orthonormal basis, as Gram Schmidt does not require $\langle A|B\rangle=\langle B|A\rangle ^*$. So at least that stuff still works out.

EDIT- I just realised that, while Gram Schmidt may not require $\langle V|V\rangle$ to be strictly positive, it does require $\langle V|V\rangle$ not to be 0 for non-zero vectors $|V\rangle$, because only then can we rescale the basis vectors by their norm to get a unit vector.

EDIT- I also realised that the Cauchy Schwarz and Triangle Inequalities would no longer make sense without this norm. Maybe these are useful results too.

psmears
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Ryder Rude
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3 Answers3

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$\newcommand{\Cpx}{\mathbf{C}}$Context suggests you're asking about inner products in a (finite-dimensional) complex vector space and asking why the usual definition imposes conjugate symmetry rather than the symmetry imposed for real inner products.

Let's look at the situation in $\Cpx$, writing $A = x + iy$ and $B = x' + iy'$ with $x$, $y$, $x'$, and $y'$ real. The "symmetric" (or complex-bilinear) definition of the inner product is $$ AB = (x + iy)(x' + iy') = (xx' - yy') + i(xy' + x'y). $$ By contrast, the "Hermitian" (or "conjugate-linear") definition is $$ A^{*}B = (x - iy)(x' + iy') = (xx' + yy') + i(xy' - x'y). $$ Suppose we're looking for a generalization of the real inner product. Which should we pick?

For the first, neither component (real or imaginary part) is the Euclidean dot product. For the second, the real part is the Euclidean dot product under the obvious identification of $\Cpx$ with the real plane. Score one for conjugate-linearity.

There's a fringe benefit to the second: The imaginary part is another friend of ours, the determinant or area form in the plane.

These observations generalize for complex vectors with $n > 1$ components, i.e., to the complex vector space $(\Cpx^{n}, +, \cdot)$: The real part of the conjugate-linear inner product is the Euclidean dot product on the real vector space $\Cpx^{n}$, and for free we pick up interesting extra structure (a skew-symmetric real-bilinear function) in the imaginary part.

Since an arbitrary finite-dimensional complex vector space is isomorphic to a complex Cartesian space, it's natural to adopt conjugate-linearity when we define an inner product on a vector space with complex scalar multiplication.

Added: In addressing why we prefer Hermitian symmetry for complex inner products, I may have missed the motivation about "What do we lose if an inner product is indefinite (but non-degenerate)?"

The edits to the question mention linear-algebraic properties we lose with an indefinite inner product. Another is "homogeneity of direction": In a Euclidean space (positive-definite inner product), for each pair of lines through the origin there exists a linear isometry carrying one line to the other.

When we have an indefinite inner product, this is no longer the case; we have homogeneity on the sets of "positive" lines, of "null" lines, and of "negative" lines. That in turn means some lines through the origin do not hit the unit sphere.

<>

This is getting a little far afield, but from my perspective as a geometer, the great "loss" in dropping positivity is not linear-algebraic but topological: In an indefinite inner product space we lose compactness of the unit sphere.

Here's a basic example of the resulting complications: Suppose $(M, g)$ is a compact manifold equipped with a Riemannian metric (positive-definite at each point). At each point, the sphere of unit vectors is compact. Consequently (because local product structure), the unit sphere bundle is compact. Now, unit-speed geodesics of $(M, g)$ may be viewed as integral curves of a unit vector field on the unit sphere bundle. Compactness of the unit sphere bundle ensures completeness of the flow: On a compact Riemannian manifold, geodesics exist for all time.

On a compact indefinite (pseudo-Riemannian) manifold, however, geodesic flow can be incomplete. Einstein Manifolds by A. Besse contains an example of a Lorentz-signature metric on an ordinary $2$-torus with this property. If the light cones (null directions of the metric) tilt suitably from point to point, there is a unit-speed timelike geodesic that in finite time winds infinitely many times around the torus.

This isn't to say mathematicians can't work with indefinite metrics, just that doing so is technically more challenging, and some properties of positive inner products do not generalize to indefinite non-degenerate inner products.

Andrew D. Hwang
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    +1 for the sentence "Score one for conjugate-linearity". – Lee Mosher Jun 13 '21 at 14:28
  • The Hermitian definition reminds me of the quaternion product and the geometric product from geometric algebra. I wonder if all these are based on the same underlying idea. The quaternion product is also the sum of dot and cross products, where the cross product is basically like a determinant. – Ryder Rude Jun 13 '21 at 16:43
  • In the sense of "conjugate before multiplying", probably yes. :) <> The business about the imaginary part of an Hermitian inner product being related to determinants is either "a thing that happens because algebra", or "some wrangling with trigonometry". It does, however, play a central role in complex geometry, where the imaginary part of an Hermitian metric is the so-called Kähler form. – Andrew D. Hwang Jun 14 '21 at 21:10
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A positive norm allows us to turn the inner product space into a metric space by measuring the distance between vectors with $d(x,y) = \|x-y\| = \sqrt{(x-y)\cdot (x-y)}$ .Lots of things rely on a notion of a distance derived from the inner-produc and this helps these concepts generalize. For instance, the concept least-squares solutions to $A\mathbf{x} = \mathbf{b}$ still make sense for complex inner product spaces and the process of solving it remains essentially the same.

Rioghasarig
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The definition of a norm includes that $\|x\|>0$ if $x\ne 0$, the additive identity. (The other two parts of the definition involve multiplication by a scalar in the field and the triangle inequality.) Since the norm is a function $\|\cdot\|$ from vectors in a vector space $S\rightarrow\mathbb R$, sometimes the norm is denoted by $\rho(x)$. If the norm isn’t strictly positive for all nonzero vectors, i.e. the condition is relaxed, the function is called a seminorm.

The norm contains the notion of length. Any inner product can induce a valid norm, if we take the inner product to be positive if $x\ne0$. Then the norm induced by the inner product is $\|x\|=\sqrt{\langle x,x\rangle}$ and satisfies the definition of a norm. If $\|\cdot\|$ is a norm then $\Delta(x,y)=\| x-y\|$ satisfies the conditions to be a metric.

Vons
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