I am including this information about real higher order derivatives because it does not seem to be common knowledge. I also include a review of the complex Hessian.

If $f:\mathbb{R}^n \to \mathbb{R}^k$ is a function, then its derivative $Df:\mathbb{R}^n \to \textrm{Hom}(\mathbb{R}^n,\mathbb{R}^k)$. $Df(\mathbf{p}):\mathbb{R}^n \to \mathbb{R}^k$ is a linear map between tangent spaces for each point $\mathbf{p} \in \mathbb{R}^n$. $Df(\mathbf{p})$ is defined as the unique linear map so that

$$ \lim_{\vec{h} \to \vec{0}} \frac{\left| f(\mathbf{p}+\vec{h})-f(\mathbf{p})-Df(\mathbf{p})(\vec{h}) \right|}{|\vec{h}|} = 0 $$

Endowing $\textrm{Hom}(\mathbb{R}^n,\mathbb{R})$ with any norm you desire (the operator norm for example), we find that we can differentiate $Df$ to obtain a map $D^2f:\mathbb{R}^n \to \textrm{Hom}(\mathbb{R}^n,\textrm{Hom}(\mathbb{R}^n,\mathbb{R}^k))$.

Fundamentally, $D^2f$ allows us to approximate changes in the derivative:

$$D^2f(\mathbf{p})(\vec{h}_1)(\vec{h}_2) \approx Df_{\mathbf{p}+\vec{h}_2}(\vec{h_1}) -Df_\mathbf{p}(\vec{h_1})$$ We can reinterpret this as $D^2f(\mathbf{p}): \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}^k$ being a bilinear map. In the special case $k=1$, $D^2f(\mathbf{p})$ is a bilinear form.

The second derivative as a bilinear form is given by the formula

$$ D^2f = \sum_{j,k=1}^n \frac{\partial^2 f}{\partial x_j \partial x_k} dx_j \otimes dx_k $$

and can be represented by a matrix called the "Hessian matrix" of $f$. Famously it turns out that this bilinear form is symmetric, and so is its corresponding matrix.

All of this generalizes easily to higher dimensional derivatives: the $D^kf(\mathbf{p})$ can be thought of as a $k$ linear form on $\mathbb{R}^n$ which measures changes in $D^{k-1}f$. The formula is just

$$ D^kf(\mathbf{p}) = \sum_{|I|=k} \frac{\partial^k f}{\partial x_I} dx_I $$

where I am using multi index notation.

For functions $f:\mathbb{C}^n \to \mathbb{C}$ the so called "Complex Hessian" of $f$ is incredibly important. It is given by the formula

$$\mathcal{H} = \sum_{j,k=1}^n \frac{\partial^2 f}{\partial z_j \partial \overline{z}_k} dz_j \otimes d\overline{z}_k$$

It turns out that another way to write this is as $H(v,w) = i\partial \overline{\partial} f(v,J(w))$ where $\partial$ and $\overline{\partial}$ are the Dolbeault operators and $J:\mathbb{C}^n \to \mathbb{C}^n$ is the complex structure map $J(z_1,z_2,...,z_n) = (iz_1,iz_2,...,iz_n)$.


  1. Does anyone have any intuitive reasons for the extreme importance of the complex Hessian in several complex variables? It is really totally fundamental to the subject, and I use it all the time, but I still do not have a completely "natural" justification for it.

  2. We can consider a function $f:\mathbb{C}^n \to \mathbb{C}$ as a complex valued function on $\mathbb{R}^{2n}$, and use the discussion above to generate a notion of a higher order derivative tensor $D^kf$, but this completely ignores the complex structure. Is there a natural generalization of the Complex hessian to higher order complex derivatives? Could this be related to the notion of "finite type" in several complex variables?

  • $\begingroup$ I don't have a good answer (I'm not an SCV guy), but (1) it has a coordinate free expression as you said (2) it specializes to (a constant times) the laplacian in 1 complex variable. $\endgroup$ – Donu Arapura Apr 14 '14 at 21:59
  • $\begingroup$ @DonuArapura Yes, these are basically the motivations I have for the Hessian as well. I would like to have a kind of "characterization" style reason for liking it. Something beginning "The complex hessian is the unique sequilinear form such that..." might be satisfying. I have such nice "reasons" for liking the Dolbeault operators: on functions they are the unique splitting of the real derivative into complex linear and complex antilinear parts. $\endgroup$ – Steven Gubkin Apr 14 '14 at 22:10
  • $\begingroup$ Hopefully having such a characterization would also point the way towards higher dimensional derivatives as well. $\endgroup$ – Steven Gubkin Apr 14 '14 at 22:11
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    $\begingroup$ I have just a remark on your comment about interpreting the complex derivative on even-dimensional real spaces. When you say that it completely ignores the complex structure, have you seen that $\mathbb C$-analyticity of a function $f:\mathbb C \to \mathbb C$ is equivalent to the real derivative of $f: \mathbb R^2 \to \mathbb R^2$ being anti-symmetric? I would imagine this type of characterization would generalize in some form to higher dimensions. $\endgroup$ – user39190 Apr 14 '14 at 22:18
  • $\begingroup$ @ntropy: yes this is more or less included in my comment above: the real derivative $Df:\mathbb{R}^{2n} \to \mathbb{C}$ splits uniquely into a complex linear part $\partial f$ and a complex antilinear part $\overline{\partial} f$. The function is holomorphic iff $\overline{\partial} f = 0$, i.e. if $Df$ was already complex linear to begin with. $\endgroup$ – Steven Gubkin Apr 14 '14 at 22:37

The reason the complex Hessian (actually, it ought to be called the 'Hermitian Hessian', since it defines an Hermitian form at every point, but 'the complex Hessian' is entrenched in the literature) is so important is that it is the unique (up to constant multiples) linear second order differential operator from real functions to $(1,1)$-forms that is invariant under all changes of holomorphic coordinates. (That it is invariant follows from the formula that you wrote down, i.e., $\mathcal{H}(f) = i\,\partial\bar\partial f$, which only uses the operators $\partial$ and $\bar\partial$ on differential forms.)

Note that in real variables, the 'real Hessian' is not invariant under all changes of real variables. In fact, if $f:\mathbb{R}^n\to\mathbb{R}$ is such that $\mathrm{d}f_p\not=0$, then one can always find local coordinates $x^i$ centered at $p\in\mathbb{R}^n$ so that $f = f(p) + x^1$, so that, in these coordinates, the Hessian of $f$ vanishes. In fact, in order to define the Hessian of $f$ at a point that is not a critical point, you need to have a connection, and the Hessian (with respect to the connection) is only well-defined up to transformations that preserve the connection. (In 'flat space', i.e., $\mathbb{R}^n$ with its standard flat connection, the Hessian is given by the formula you wrote.)

Added note on the second question (1 December 2016): I forgot about this anwer until someone upvoted it again today, and that reminded me that I had not addressed the second question at all, so I thought I'd make a remark about it now: There is no natural notion of a 'complex third derivative' of a real-valued function on a complex manifold because it turns out that the complex biholomorphisms act with some open orbits on the $3$-jets of functions, and those open orbits are in one-to-one correspondence with its open orbits on the $2$-jets of functions (which correspond to the fuctions $f$ for which $\mathcal{H}(f)$ is nondegenerate as a $2$-form).

However, when you to go $4$-jets, the story changes. There is a natural notion of a $4$th-order operator, but it is nonlinear. To see this in a simple example, consider the case of one complex dimension: Building on the second order linear operator, expressed in terms of a local holomorphic coordinate $z$ as $$ \mathcal{H}(f) = i\,\partial\bar\partial f = i\,f_{z\bar z}\,\mathrm{d}z \wedge \mathrm{d}{\bar z}, $$ we have the $4$th-order quadratic operator $$ \begin{aligned} \mathcal{K}(f) &= i\,\partial\bar\partial \log(f_{z\bar z})\otimes \mathcal{H}(f)\otimes \mathcal{H}(f)\\ &= -i\,\bigl(f_{z\bar z}\,f_{zz\bar z\bar z} - f_{zz\bar z}\,f_{z\bar z\bar z}\bigr)\,(\mathrm{d}z \wedge \mathrm{d}{\bar z})^{\otimes 3}, \end{aligned} $$ which (essentially) gives the curvature of the Hermitian form $\mathcal{H}(f)$ when the latter is a positive $(1,1)$-form, suitably shifted as a tensor to remove the singularities that would show up where $\mathcal{H}(f)$ vanishes.

Of course in higher dimensions, the corresponding $4$th-order operator is just a version of the Kähler curvature tensor. There's no need to write it out here.

  • $\begingroup$ Thanks for this answer. Any insight into the second question? $\endgroup$ – Steven Gubkin Apr 14 '14 at 23:41
  • $\begingroup$ Is there perhaps a 4th order differential operator from functions to (2,2) forms which is invariant under homolomorphic changes in coordinates? I guess I should play with trying to write some down... $\endgroup$ – Steven Gubkin Apr 14 '14 at 23:45

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