# What is a complex inner product space “really”?

This is an extended re-post of a question that I have asked on MSE not a long time ago. But anyway, it seems more appropriate for MO.

To begin with, in a real inner product space we have a geometric intuition for the inner product. In the finite dimensional case, the inner product of two vectors is the product of their lengths (norms) times the cosine of the angle between them. Reverse engineering this suggests that the purely algebraic properties of an abstract inner product give it the properties of the "length of the projection" (scaled somehow). In particular, we can think of two vectors with zero inner product as orthogonal geometrically (and even call them that way) and bring all the geometric notions related to orthogonality to the abstract, possibley infinite dimensional case (with the appropriate care and restrictions of course). My question is, what is the geometric or otherwise intuition behind the abstract notion of a complex inner product space?

Here are a few thoughts:

1) This is a good mathematical structure to model some physical phenomena. An (if not the) example is quantum mechanics. This is an interesting line of thought. One problem is that I don't know enough quantum mechanics to follow it more deeply. If it is possible to explain in an elementary as possible way, What does it mean that a state of a particle is an element of a complex Hilbert space I would very much want to hear about it. I would also like to hear about other, hopefully more elementary, phenomena modeled by complex inner spaces. In particular, ia there any such phenomenon modeled by a finite-dimensional complex inner product space?

2) This is a good mathematical tool for other mathematical theories. Perhaps, unitary representations of groups or of other algebraic structures. Again, I don't have enough background in representation theory myself. explicit examples of such utility are welcome.

3) it is a good tool to investigate real structures by complexification and exploit of the good properies of the complex field (such as being algebraically closed). There are a lot of such examples in linear algebra, such as the classification of orthogonal maps, but I haven't seen such examples in the context of inner product spaces.

I would like to stress that saying that this is somehow algebraically natural analogue of real inner product spaces and that it has a lot of nice properties is somehow not enough in my opinion. Also not very satisfying is saying that it has application in such and such very advanced theories without elaboration.

Thanks!

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Is your question about the actual datum of an inner product, or are you interested in the whole structure of an inner product space, so that one could substitute the inner product by a norm satisfying the parallelogram law (this description is valid for both the real and the complex case), which is (in my opinion) much closer to geometric intuition? – user2035 Nov 3 '11 at 9:23
I wonder why was it downvoted... @a-fortiori, I don't mind using an equivalent definition if it helps to think about this things. My feeling is that if you can really "explain" one of the definition then it should be possible to say something about the other too, but maybe I am wrong. – KotelKanim Nov 3 '11 at 16:48
I didn't downvote, but I would vote to close if I didn't have superpowers, because your question seems to be both basic and ill-focused. The real part of a complex inner product is a real inner product on the underlying real vector space, so you get all the angles, lengths, etc. you see in real geometry - this is much stronger than a "natural analogue". The complex structure on the vector space gives you additional information that is reflected in the imaginary part of the inner product. By the way, in QM, it is the states of particles, not just particles, that are lines in Hilbert space. – S. Carnahan Nov 3 '11 at 17:13
I respect your opinion. This is perhaps a basic and ill-focused question that should be closed, yet I would at least like to understand why (and maybe it is still not a legit thing to ask). I know that the real part of the inner product is a real inner product and I should have addressed that in the formulation of the question, but the main point is what exactly is the "additional structure" the imaginary part provides. About QM, I will edit the question, though it is obviously not the main thing that bothers you. – KotelKanim Nov 3 '11 at 20:44
I think it's phrasing: I would shorten it a lot, say very briefly stuff you know already (or more at the end), so to focus on the parts that should be "interesting to mathematicians." People can be too quick to assume the question is too basic. – Elizabeth S. Q. Goodman Nov 4 '11 at 7:00

I'm not sure if this is what you're looking for. But to me, a Hermitian metric is just $g+i\omega$, where $g$ is a real inner product, and $\omega$ is a symplectic form (alternating, but still non-degenerate).

To share my simplest intuition, once you believe that this concept is useful: $g$ tells you how a pair of vectors measure up geometrically in $\mathbb R^{2n}\simeq\mathbb C^n$, as you've already noted. But $i\omega$ tells you how much closer to linearly dependent the vectors are now that complex scalars are allowed, so it's still a lot like measuring angle. That is, if $v,w$ are orthonormal in the real sense, then if $i\omega(v, w)=i$, they are in the same complex line; if $i\omega(v, w)=0$, they are just as orthogonal in the complex sense as in the real sense.

It may be just the fact that I do symplectic geometry that makes me think of $\omega$ as such useful geometric information, but once you start looking there are many settings that can be made symplectic. Probably the most well-known one is the symplectic geometry of a cotangent bundle as a setting for Hamiltonian methcanics: there are coordinates for position and velocity (er, momentum), and the complex structure, measured by the metric and the symplectic form, tells you how they're related.

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This is indeed an interesting perspective. Thank you! – KotelKanim Nov 4 '11 at 7:41

A complex inner product space is just a real inner product space along with a designated 90 degree rotation (where this means a linear operator sending every vector to an equally large but perpendicular vector)${}^*$. If you understand why one might care about real inner product spaces, and you understand why one might be furthermore interested in 90 degree rotations (or equivalently, rotations by constant angles more generally${}^+$), then you understand why one might care about complex inner product spaces.

[${}^*$: Specifically, given such a real inner product space $V$ with 90 degree rotation operator $J$, view $V$ as a complex vector space by taking $(a + bi)v = (a + bJ)v$, of course. And define a complex inner product $*$ from the real inner product $\cdot$ via $v * w = v \cdot w + (v \cdot Jw)i$. To go in the other direction, from a complex inner product space to a real inner product space with 90 degree rotation operator, is even easier: take $v \cdot w$ to be the real component of $v * w$, and take $Jv$ to be $iv$. One can straightforwardly check that these two processes are inverse to each other and produce structures satisfying the appropriate axioms.]

[${}^+$: Let us say $R$ is a rotation by a constant angle if $R$ is a linear, length-preserving operator such that $\frac{Rv \cdot v}{v \cdot v}$ is constant. It is straightforward to show that this is equivalent to $R$ being of the form $\cos(\theta) + \sin(\theta)J$ for some angle $\theta$ and some 90 degree rotation operator $J$, so the study of constant angle rotations in general reduces to the study of 90 degree rotations in particular]

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This is great (closest to my intentions so far), but I would like to ask you for some clarifications if possible. First, When you say "a 90 degree rotation", do you simply mean an orthogonal linear map $J$ such that $<v,Jv>=0$ for all $v\in V$? and why one might be interested in such rotations? because it gives you a "canonical" orthogonal vector $Jv$ for each $v$ and a canonical "direction" of rotation? can you be more specific about why is it a useful structure? (I can think of some things myself, but I would like to hear more) – KotelKanim Nov 4 '11 at 12:12
BTW, it seems related to Elizbeth's answer regarding the connection to symplectic geometry. Sadly, this is yet another area I am not familiar with more than just the basic definitions. – KotelKanim Nov 4 '11 at 12:14
Yes, this is related to my answer. The "90 degree rotation" means i, itself; we call that J, a "complex structure", it allows you to define rotation in general, and by the way that means you have an orientation too. But in general J just has to square to -1, it doesn't have to be compatible with the metric (that is, doesn't have to be 90 degrees, or length-preserving). But here's a fun exercise: if you have 2 out of 3: $g, J, \omega$ and they're compatible (try out what that could mean in each pair!), then you get the third one for free. – Elizabeth S. Q. Goodman Nov 4 '11 at 17:40
Here is why I said $J$ has to be length preserving: Suppose you have a real inner product $\cdot$, linear operator $J$, and complex inner product $\*$, related by the correspondence in my post. Then $Jv \cdot Jv$ is the real component of $(iv) \* (iv)$, which by "sesquilinearity" is the real component of $v \* v$, which is $v \cdot v$. Thus, $Jv \cdot Jv = v \cdot v$. – Sridhar Ramesh Nov 4 '11 at 18:54
Similarly, here is why I said $J$ has to rotate by 90 degrees: $Jv \cdot v$ is the real component of $(iv) \* v$, which by "sesquilinearity" is the real component of $i(v \* v)$, which by positive-definiteness is $0$. Thus, $Jv \cdot v = 0$. It follows from those properties that $J^2 = -1$ [consider $(x + Jx) \cdot J(x + Jx)$], but just taking $J$ to be an arbitrary linear operator satisfying $J^2 = -1$ will not, under the transformation outlined above, produce a complex inner product (with "sesquilinearity" and positive-definiteness). – Sridhar Ramesh Nov 4 '11 at 19:06

This is a purely algebraic answer, so maybe not what you're looking for.

What would an inner product on a left $A$-module $M$ over a noncommutative algebra $A$ be? It should give a map $M \to M^*$ where $M^* = Hom_A(M,A)$, but that's a right $A$-module. We can teach it to be a left $A$-module again, if we have a chosen anti-automorphism $i:A\to A$. Eventually we'd like the natural function $M \to M^{**}$ to be $A$-linear, probably, so we should also ask that $i$ be an involution.

If $A = \mathbb C$ then there are two (continuous) such $i$: the identity, and the complex conjugate. I claim that the latter is the more natural one, or at least that this isn't crazy. If you think of $\mathbb C = \left\{\begin{pmatrix} a&b\\ -b&a \end{pmatrix} : a,b\in \mathbb R\right\}$, then (1) it's weird that $\mathbb C$ is commutative, since matrices shouldn't commute, and (2) $i$ is transpose.

If you go up to $A = \mathbb H$ then you no longer have this weird choice of $i =$ identity available. But transpose still works (thinking of $A$ as certain $4\times 4$ real matrices).

Anyway, if you now unwind the $A$-linearity of $M \to M^*$, where $A=\mathbb C$ and $i(z) = \overline{z}$, you get sesquilinearity.

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Suppose first that you're not interested in the complex situation at all and just consider real inner product spaces (with positive definite inner product for simplicity). Then you have the notion of transposed operators with respect to the inner product and you get a notion of normal operators this way. Now the spectral theory for normal operators is very nice provided they are even symmetric ones: they can be diagonalized. Now there are other normal operators as e.g. the rotations where $D D^T = 1 = D^T D$. However, they do not even need to have an eigenvector (depending on the dimension...) So their normal form is much more complicated and cumbersome for many reasons.

The reason is that the characteristic polynomial needs not to have real roots. But as we know, over the complex numbers we have the fundamental theorem of algebra telling that it decomposes into linear factors. From this one gets the idea that complexifing the real vector space to a complex one might be a good idea for spectral purposes. Indeed, doing so allows now to diagonalize rotations, they have eigenvalues on the unit circle.

Of course, one still has to see how "normal" is translated into the complex situation: you also have to complexify the inner product leading to a sesquilinear inner product (not a complex bilinear one, this is sort of useless here).

Then of course, you loose the interpretation of the inner product encoding an "angle" between vectors, but you still have a Cauchy-Schwarz inequality. The only thing which remains after this complexification is that zero remains zero: so it still makes sense to talk about orthogonal vectors.

Well, this might be a motivation why complex Hilbert spaces arise quite naturally, even if you're only interested in real ones, supporting your point 3.)

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I think it's important to note what translating the concept of "normal operator" to the complex setting yields. The point is that any operator can be decomposed into real and imaginary parts, which are self-adjoint and hence diagonalizable. It's easy to check that an operator is normal if and only if its real and imaginary parts commute, so that they are simultaneously diagonalizable. – MTS Aug 7 '13 at 22:06

The geometric significance of a complex inner product space is that it is one where the orthogonality relation is symmetric: if $v$ is perpendicular to $w$ then $w$ is perpendicular to $v$.

Background

1. Theorem: Let $V$ be a vector space (over any field, not necessarily finite dimensional) and $B$ be a bilinear form on $V$. We set $v \perp w$ in $V$ when $B(v,w) = 0$. This orthogonality relation on $V$ is symmetric if and only if the bilinear form $B$ is symmetric (meaning $B(v,w) = B(w,v)$ for all $v$ and $w$) or alternating (meaning $B(v,v) = 0$ for all $v$, which outside of characteristic 2 is the same as $B(v,w) = -B(w,v)$ for all $v$).

2. If you look at the proof that $n \times n$ orthogonal matrices have real eigenvalues, the proof works in ${\mathbf C}^n$ with its "complex" inner product, where scaling in one of the coordinates can be pulled out using complex conjugation. That is how you show an eigenvalue satisfies $\lambda = \overline{\lambda}$. I think this is a pretty good basic motivation for believing that complex inner products can be useful creatures. You wouldn't be able to get such a conclusion with ${\mathbf C}^n$ and its naive inner product (not using complex conjugation on one of the coordinates).

3. Theorem: Let $V$ be a complex vector space with dimension at least 2 (not necessarily finite dimensional) and $B$ be a non-degenerate sesquilinear form on $V$. Set $v \perp w$ in $V$ when $B(v,w) = 0$. This orthogonality relation on $V$ is symmetric if and only if there is some $\alpha \in {\mathbf C}^\times$ such that $\alpha{B(v,w)} = \overline{\alpha{B(w,v)}}$ for all $v$ and $w$.

In item 3, the function $\alpha{B}$ at the end is sesquilinear and has the same orthogonality relation as $B$ does, so effectively this theorem says that up to scaling $B$ by a nonzero constant the only way the condition $B(v,w) = 0$ is symmetric in $v$ and $w$ is when $B$ is Hermitian, i.e., $B(v,w) = \overline{B(w,v)}$ for all $v$ and $w$.

(Of course ${\mathbf C}^n$ with its naive inner product has a symmetric relation of perpendicularity, but that naive inner product is bilinear, not sesquilinear.)

The proofs of the two theorems mentioned above are completely elementary but involve tedious calculations. See Artin's "Geometric Algebra", pp. 113--114.

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You can view complex inner product space as an object that encompasses a set of real inner product spaces which are not necessarily positive definite. I.e. from $\mathbb{C}^n$ with the standard inner product you can get Euclidean real space as well as Lorentzian by choosing appropriate real form. So if you want to find a real geometric intuition for complex inner product spaces (assuming there is one!) you should probably better start with inner product spaces of indefinite signature.

As an example of use of inner product spaces I suggest the theory of simple Lie algebras. The theory is developed in the complex case and one has a natural (invariant) inner product there - the Killing form. From one complex Lie algebra you can get several real Lie algebras and their properties (i.e. whether they are compact or not) are determined precisely by the signature of the restriction of Killing form.

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This answer is not complete, but I think the points are important.

A complex vector space is one in which every vector $v$ has a special, oriented plane going through it, where each $\alpha \in {\mathbb C}$ rests upon the vector $\alpha v$. With this interpretation it is easy to see why $(\alpha + \beta)v = \alpha v + \beta v$, but actually a bit tricky to see why $\alpha(v + w) = \alpha v + \alpha w$.

A real inner product gives a vector space a notion of geometry in which angles can be computed. In a normed complex vector space (which suffices to provide a notion of distance), the geometry of each plane $\alpha v$ going through $v$ looks like a rescaling of the usual complex plane because $|(\alpha v - \beta v)| = |\alpha - \beta| |v|$.

One basic thing that separates an inner product space (or Hilbert space, rather) from a normed vector space is the fact that for any (closed) subspace $W$ of $V$ the projection $P_W(v)$ onto $W$, which finds the unique element in $W$ of minimal distance to $v$, turns out to be a (complex!) linear map. That's something very special about the shape of ellipsoids (unit balls in inner product spaces) which is not common to other kinds of unit "balls" -- for other shapes of the unit ball, this minimizer won't even be unique. This fact about linear projections, when applied to the space $W = \{ \alpha v \}$ can explain the bilinearity of the inner product. (I think the real analogue of this fact can be demonstrated, or at least interpreted, by ruler and compass construction.) For Hermitian inner product spaces, we need a complex normed vector space, so we also want the rotations $e^{i \theta} v$ to be length-preserving, so that's one explanation for why we want conjugate linearity in the second variable of a Hermitian inner product.

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