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\newcommand{\bbc}{\mathbb{C}}
\newcommand{\Int}{\operatorname{Int}}
\)Navigation: 1. Exact, conservative and closed forms | 2. Locally exact forms and singular homology | 3. Applications to complex analysis

Preliminaries

Let \(F\) be a complex Banach space, let \(U\subseteq\bbc\) be an open set, and let \(f:U\to F\). Recall that the complex Fréchet derivative of \(f\) at \(z\in U\), if it exists, is a \(\bbc\)-linear map \(Df(z):\bbc\to F\). Not all real differentiable functions on \(\bbc\) are complex differentiable: for example, the (real) derivative of \(f(z)=\overline{z}\) (i.e. \(f(x+yi)=x-yi\)) at any point is represented by the matrix $$\begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix},$$ but \(Df(z)\) is clearly not a \(\bbc\)-linear map.

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Navigation: 1. Exact, conservative and closed forms | 2. Locally exact forms and singular homology | 3. Applications to complex analysis

As in part 1, \(E,F\) are Banach spaces and \(U\subseteq E\) is an open set. Recall that a form \(\omega:U\to L(E,F)\) is exact if \(\omega=Df\) for some \(f:U\to F\), closed if it is differentiable and \(D\omega(x)\in L(E,E;F)\) is symmetric for every \(x\in U\), and locally exact if it is closed and for every \(x\in U\) there is a neighborhood \(V\subseteq U\) of \(x\) on which \(\omega\) is exact. Last time we showed that every closed \(C^1\) form is locally exact, and that every closed form on an open subset of \(\mathbb{R}^2\) is locally exact (even if is differentiable but not \(C^1\)).

So far, we have only defined line integrals along \(C^1\) paths and curves. It turns out that for locally exact forms, we can extend the definition to paths that are merely continuous (and not necessarily differentiable).

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Navigation: 1. Exact, conservative and closed forms | 2. Locally exact forms and singular homology | 3. Applications to complex analysis

The line integral is a useful tool for working with vector fields on \(\mathbb{R}^n\), (co)vector fields on manifolds, and complex differentiable functions. However, it is often unclear how these different versions of the line integral are related to each other. In the next few posts, I will be presenting a very general form of the line integral along with the standard theorems and some basic applications. Familiarity with the Fréchet derivative is assumed.

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Last time we derived a formula for the derivative of the matrix exponential. Here we will be focusing instead on the expression $$D\exp(x)u=\exp(x)u=u\exp(x),$$ which holds whenever \(u\) commutes with \(x\). In this post, \(E\) denotes a real Banach space and \(L(E)\) denotes the space of continuous linear operators on \(E\).

ODEs

The (matrix) exponential can be used to solve certain types of first-order linear systems of ordinary differential equations with non-constant coefficients: not only can we solve \begin{align}x'(t)&=x(t)+y(t) \\ y'(t)&=x(t)+y(t),\end{align} but we can also solve \begin{align}x'(t)&=tx(t)+y(t) \\ y'(t)&=x(t)+ty(t).\end{align}

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$$
D\exp(x)u = \int_0^1 e^{sx}ue^{(1-s)x}\,ds.
$$ This intriguing formula expresses the derivative of the exponential map on a Banach algebra as an integral. In particular, using “matrix calculus” notation we have the formula $$
d\exp(X)= \int_0^1 e^{sX}(dX)e^{(1-s)X}\,ds
$$ when \(X\) is a square matrix. As we’ll see, this is not too hard to prove.

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