

Complex Analysis Question

I have a problem with the way complex analysis is usually taught and studied. The main objects there seem to be the holomorphic functions, characterized by the fact that they satisfy the Cauchy–Riemann equations. Now I read that satisfying this pair of equations is the same as the Jacobian matrix having the form of the matrix representation of a complex number. So a holomorphic function is essentially just a function whose Jacobian is also a complex function, not some general matrix.

My question is, why do this? When we study real functions of n variables, we simply take the Jacobian of that function and consider it to be the "derivative". We do not restrict ourselves to the functions whose Jacobians are diagonal. That would be nonsensical. Sure, we would get some nice results, but this arbitrary choice would be crippling to our understanding in the long run.

So why this seemingly unnecessary restriction? Wouldn't complex analysis be a much richer subject if we would consider the "derivatives" of "differentiable" complex functions? If that were the case, we would simply clarify that all the nice results in complex analysis apply only to the holomorphic functions, while the rest of it is just as messy as real analysis.

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level 1
22 points · 5 years ago

Suppose you wanted to study complex manifolds: these should be manifolds which are locally modeled on Cn rather than Rn. What does it mean for a map f: X->Y of complex manifolds to be smooth?

You can pretend that Cn = R2n and just talk about the derivative df in the usual sense, but it's generally very important for analysis that the derivative be a linear map between tangent spaces. For real manifolds this is nothing new, because you've been thinking of linear maps over R, but for complex manifolds you'd like the differential df to be C-linear rather than just R-linear. In other words, X comes with a "complex structure" J, which is a linear map TX -> TX satisfying J2=-1 and some "integrability" condition, and likewise for Y; if we took X to be C, then each tangent plane is also a copy of C and J is the usual multiplication by i. Now df, maybe viewed as a 2n x 2n real matrix, is only complex linear if df(av) = a*df(v) for any complex a and tangent vector v. Equivalently, we want df to commute with J (the analogue of the complex number i), meaning that df*J = J*df. In the case of maps C->C, with Jv = i*v, this is exactly the Cauchy-Riemann equations.

As for why we would do this, one great motivation is algebraic geometry. It's much easier to do algebraic geometry over C than over R, because C is algebraically closed and R is not. Smooth, irreducible complex varieties are also complex manifolds, and the complex structure tells you a lot about the topology (see for example Hartogs' theorem or Hodge theory or Dolbeault cohomology), so it's natural to focus your attention on the phenomena which occur when you restrict your attention to holomorphic functions. You can just stick to real analysis if you prefer, but then all of these phenomena disappear and you're no closer to understanding something like complex algebraic geometry than you were before.

This also lets you understand a lot about things which aren't quite complex, like symplectic manifolds. If a manifold has a symplectic structure (which occurs naturally throughout physics, e.g. in Hamiltonian mechanics), then you can find a compatible "almost complex" structure J which is basically a complex structure without the integrability. If you have, say, a Riemann surface S and a symplectic manifold X, you can still write down the Cauchy-Riemann equations for a map S->X and call solutions to these equations "pseudoholomorphic curves", and you can learn a lot about your symplectic manifold by understanding such maps, imitating a whole lot of the interesting parts of both complex analysis and algebraic geometry along the way. These techniques are powerful enough that almost all progress in symplectic geometry these days uses them in one form or another.

level 2
Original Poster5 points · 5 years ago

The third and fourth paragraph of your comment did more than just answered my question. Thank you for that. The things in the second paragraph are definitely new to me, and I will look into them.

level 2
Representation Theory3 points · 5 years ago

Another remark:

/u/mathman1, what sequence of books would you recommend for learning in this area? Obviously Griffiths-Harris, but both as good prereqs for that book and as books beyond (or instead of) G-H?

level 3
5 points · 5 years ago

I would start with a good book on Riemann surfaces -- there are some recommendations on MO here, and I originally learned from Griffiths's "Introduction to Algebraic Curves" but I know Miranda's book is also good and I can't say how the others compare. I'm not an algebraic geometer so I can't make the best recommendations for something after G-H, but maybe Voisin's "Hodge theory and complex algebraic geometry"?

If you want to learn things from a more complex analytic perspective, there's a book by Hörmander but I would recommend starting from the MIT OpenCourseWare notes for 18.117, "Topics in several complex variables." The whole course is great, but I think the material in lectures 7-30 are especially important. After this you could try an introduction to symplectic geometry, such as the book by Cannas da Silva or the two volumes by McDuff and Salamon ("Introduction to symplectic topology," then "J-holomorphic curves and symplectic topology").

level 2
Representation Theory1 point · 5 years ago

This is most definitely the right answer.

level 2

that was brilliant, thank you. This is a new perspective on the C-R equations for me.

level 1
Arithmetic Geometry10 points · 5 years ago

I think the reason is that holomorphic functions are the functions that are complex-differentiable, that is, such that the limit of [f(z) – f(w)]/(z – w) as z → w exists for all w in the domain. This is the natural generalization of differentiability to the complex numbers, and consequently, holomorphic functions "respect the complex structure" in some sense.

If, on the other hand, we looked at all functions that are real-differentiable, not just the complex-differentiable ones, then the complex structure becomes irrelevant; it's the same as doing analysis on R2n. So, the restriction to holomorphic functions is because without that restriction, we get something already covered by real analysis.

level 2
Original Poster2 points · 5 years ago

I may be gravely wrong, but are we not essentially doing real analysis in R2n with the functions that have diagonal Jacobians only? These functions share many nice properties with the holomorphic functions anyway.

If we are doing nothing new to begin with, then why shouldn't we generalize further and keep doing old real analysis, just in a new setting?

level 3
Arithmetic Geometry9 points · 5 years ago

When you generalize, you sometimes lose nice theorems. Complex analysis is the study of the additional structure and properties that arise specifically among functions that respect the complex structure. These theorems mostly aren't true in the context of real analysis on R2n, so by trying to gain generality, you'd actually miss an entire theory. pun intended.

Similarly, if you try to "generalize" the theory of manifolds by looking at, say, arbitrary Hausdorff spaces, then you lose much of the structure that makes manifolds interesting. Mathematics isn't only about finding more general objects to study, but also about working out the more detailed structure of narrower, more specific objects.

On the other hand, if you can find some natural class of real functions that behave analogously to holomorphic functions, that might be an interesting topic in its own right. But in that case, it'd be interesting because it mimics complex analysis in a different context.

level 4
Original Poster-2 points · 5 years ago

When you generalize, you sometimes lose nice theorems.

I simply can't accept that offhand. Take, for instance, conservative vector fields. There is some beautiful stuff there, no doubt. They have path independent integrals, much like the case in complex analysis. Does that mean that we shouldn't consider arbitrary vector fields? Of course not. There is much more knowledge, beauty and usefulness to be gathered out there.

level 5
Symplectic Topology6 points · 5 years ago

What he said isn't something you can accept or not, it's a simple fact. When you generalize, you consider more objects at once, and so facts that were true for a special class of such objects might not stay true for all the new objects. For instance, to use the example that you gave, it's true that every conservative vector field is irrotational, but it's not true of every vector field. You lost this fact by generalizing.

level 6
Original Poster-2 points · 5 years ago

You lost this fact by generalizing.

You lost the fact but you accommodated it inside a larger, more beautiful and comprehensive theory. Nothing wrong with that.

level 7
Symplectic Topology7 points · 5 years ago

Sure, but the point is that it's usually impossible to understand all the big theory at once. Thus, motivated by the different classes of examples we find, we separate the theory into various special cases that we can understand better.

level 7

you are the only one saying anything is wrong

level 5
2 points · 5 years ago

I simply can't accept that offhand.

More objects mean weaker assumptions, leading to weaker results. It's not a critique of the beauty or applications, it's just a fact that you can't get theorems which are quite as strong.

level 5
Algebraic Topology1 point · 5 years ago

This point of view doesn't really get you anywhere. Imagine if everyone refused to study functions because, after all, they're really just subsets of relations, which are much more general. Then there really wouldn't be any calculus to begin with.

level 6
Original Poster1 point · 5 years ago

Then there really wouldn't be any calculus to begin with.

Calculus would exist. We would just say that it is the study of functions, of that particular subset of all relations. Other relations would be studied by utilizing other theories.

Studying the general and studying its parts are not mutually exclusive. We could study complex analysis generally and conclude that this generalization is not useful after all. What baffles me is that no one has even tried doing that.

level 1

Then you'd just be doing real analysis over R2 , which is fine, but not the point.

level 2
Original Poster-1 points · 5 years ago

This way we seem to be doing real analysis with functions that have diagonal Jacobians only. I don't think that is the point either.

level 3
Representation Theory2 points · 5 years ago

If we have functions from Cn to Cm, we don't require that the jacobians be diagonal. We just require that they are mxn matrices with complex entries.

So if we have a function from C1 to C1, it's (complex) jacobian should be a 1x1 matrix with complex entries. That is, a complex number.

level 1
Geometry & Topology3 points · 5 years ago

To add to what fattybake has already said, if you assume a function is complex differentiable then it necessarily satisfies the Cauchy-Riemann equations. You can derive this yourself if you haven't seen it before. Just assume a complex function has a derivative at a point, and evaluate its partial derivatives along orthogonal paths. Matching real components and with real, and imaginary with imaginary you see that this function must satisfy Cauchy-Riemann.

As you mentioned, we seek functions whose Jacobians are the matrix representation of a complex number. It's just nice that the real and complex functions have this nice symmetry (holomorphic functions are composites of harmonic functions!). Why do we do this? So that derivatives remain within the realm of complex analysis. It's a bit like asking why don't we restrict real functions to just the natural numbers after differentiating. We could, but then it wouldn't tell us much about the reals or their functions.

level 2
Original Poster-1 points · 5 years ago


This doesn't justify the equations. Suppose we were doing real analysis over a function in two variables. Would we insist that the Jacobian matrix have the form of the matrix representation of a real number (i.e. a diagonal matrix), just so that our derivatives remain within the realm of real analysis? Of course not. That would be as pointless as it would be arbitrary.

level 3
Symplectic Topology3 points · 5 years ago

The point is that the matrix representation of a complex-differentiable map f:C -> C, viewed as a 2x2 real matrix, must satisfy the equations. It's not something that we impose arbitrarily. It's a consequence of the complex-linearity of the derivative. More precisely: f is-complex differentiable at z iff it is real differentiable at z and the real derivative Df(z) is complex linear. If you work it out, you will see that Df(z) = {{a,b},{c,d}} is complex linear iff a = d and b = -c.

level 1
Comment deleted5 years ago
level 2

Your second sentence is a common misconception. If a function is complex differentiable, then it satisfies the Cauchy-Riemann equations, but satisfying the Cauchy-Riemann equations is not sufficient for complex differentiability. The function's real and imaginary parts must have continuous partial derivatives.

level 3
Symplectic Topology3 points · 5 years ago

Actually, f need only be continuous and possess partial derivatives almost everywhere (they need not be continuous). It's the Looman–Menchoff theorem.

level 2
Original Poster0 points · 5 years ago

I understand that the satisfying the Cauchy Riemann equation is the definition of complex differentiability. However, it seems that this condition is to stringent.

Consider the class of complex functions that have an associated Jacobian. A function satisfies these equations if and only if its Jacobian is the matrix representation of a complex function.

The question is, why consider only the Jacobians for which this is the case?

Edit: I too fell into the misconception pointed out by jonah214. This comment should not read "if and only if".

level 3

it seems that this condition is to stringent

It could be that a condition is too stringent, so that the theory doesn't end up being interesting. But it turns out that complex analysis hits a kind of sweet spot of richness, being rigid enough to prove very strong theorems and general enough to be extremely useful.

level 1

Without this restriction wouldn't complex analysis just becomes the analysis of R2? You already studied that in real analysis. The whole point of complex differentiability is that the derivative is characterized by a complex number.

Edit: protocol_7 beat me to it.

level 1

I recommend you to take a look at Visual Complex Analysis in particular the chapter on differentiation. In the first sections he explains the rationale for this restriction.

level 2
Original Poster1 point · 5 years ago

Reading that chapter was one of my main motivations for asking this question. How the book explained complex differentiation and Jacobians was breathtaking for me. It was then that I thought, why be so restrictive in our definitions?

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