Existence of harmonic conjugates on simply connected domains via de Rham cohomology

Let $\mathbb{D}\subseteq \mathbb{C}$$\mathbb{D}\subseteq\mathbb C$ be a simply connected domain. Let $u:\mathbb{D}\to \mathbb{R}$$u:\mathbb D\to\mathbb R$ be a harmonic function, then there exists another $v:\mathbb{D}\to \mathbb{R}$$v:\mathbb D\to\mathbb R$ such that $u+iv$$u+iv$ is a holomorphic function.

Proof.

Consider $\ast du=\ast (u_{x}dx+u_{y}dy)=-u_{y}dx+u_{x}dy$$*d u=*(u_xdx+u_ydy)=-u_ydx+u_x dy$.

This is a closed form since $d(-u_{y}dx+u_{x}dy)=-u_{yy}dy\wedge dx+u_{xx}dx\wedge dy=\mathrm{\Delta }udx\wedge dy$$d(-u_ydx+u_x dy)=-u_{yy} dy\wedge dx+u_{xx} dx\wedge dy=\Delta u\, dx\wedge dy$. But since $u$$u$ is harmonic, $d\ast du=0$$d*du=0$.

Thus $\ast du$$*du$ is closed. Since ${\pi }_1(\mathbb{D})=0$$\pi_1(\mathbb D)=0$, we have $H_{\mathrm{d}}^1(\mathbb{D})=0$$H^1_{\mathrm{d}}(\mathbb D)=0$. Hence $\ast du$$*du$ is exact, so there exists a $v$$v$ such that $dv=\ast du$$dv=*du$, i.e.

$$v_{x}dx+v_{y}dy=-u_{y}dx+u_{x}dy⟹v_x=-u_y,v_y=u_x.$$

That is the Cauchy-Riemann condition. $◻$$\square$

 

How to understand matrix partitioning and matrix multiplication（in any Abelian Category)

Consider an Abelian category $\mathcal{A}$$\mathcal{A}$, for example the category of sheaves of abelian groups on a site, or the category of vector spaces (sheaves on a one-point space), or the category of condensed abelian groups.

Take three objects

$$X=\underset{i=1}{\overset{n}{⨁}}{X}_i,Y=\underset{j=1}{\overset{m}{⨁}}{Y}_j,Z=\underset{s=1}{\overset{k}{⨁}}{Z}_s.$$

Then

$${\mathrm{Hom}}_{\mathcal{A}}(X,Y)\cong {\mathrm{Hom}}_{\mathcal{A}}(\underset{i=1}{\overset{n}{⨁}}{X}_i,\underset{j=1}{\overset{m}{⨁}}{Y}_j)\cong \underset{i=1}{\overset{n}{⨁}}\underset{j=1}{\overset{m}{⨁}}{\mathrm{Hom}}_{\mathcal{A}}(X_i,Y_j).$$

Explicitly, the isomorphism is given by

$$f⟼\sum _{i,j}{\iota }_j\circ f_{j,i}\circ {\pi }_i,$$

where $\iota$$\iota$ and $\pi$$\pi$ are the inclusion and projection maps, respectively.

More concretely, set $e_i={\iota }_i\circ {\pi }_i$$e_i = \iota_i \circ \pi_i$; then $\mathrm{id}=\sum _i{e}_i$$\operatorname{id} = \sum_i e_i$. Hence

$$f={\mathrm{id}}_Y\circ f\circ {\mathrm{id}}_X=\sum _{j=1}^m{e}_j\circ f\circ \sum _{i=1}^n{e}_i=\sum _{i,j}{e}_j\circ f\circ e_i.$$

Now define $f_{j,i}={\pi }_j\circ f\circ {\iota }_i$$f_{j,i} = \pi_j \circ f \circ \iota_i$. Then

$$e_j\circ f\circ e_i={\iota }_j\circ f_{i,j}\circ {\pi }_i.$$

Writing $\sum _{i,j}{\iota }_j\circ f_{i,j}\circ {\pi }_i$$\sum_{i,j} \iota_j \circ f_{i,j} \circ \pi_i$ as a matrix gives

$$\left(\begin{array}{cccc}{f}_{11}& f_{12}& \cdots & f_{1n}\\ f_{21}& f_{22}& \cdots & f_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ f_{m1}& f_{m2}& \cdots & f_{mn}\end{array}\right).$$

Clearly addition of morphisms corresponds to addition of matrices.

If we also have $Z\cong \underset{s=1}{\overset{k}{⨁}}{Z}_s$$Z \cong \bigoplus_{s=1}^{k} Z_s$ and a morphism $g\in {\mathrm{Hom}}_{\mathcal{A}}(Y,Z)$$g \in \operatorname{Hom}_{\mathcal{A}}(Y,Z)$, then composition of morphisms corresponds to matrix multiplication.

If you are curious about what happens on the level of Hom groups: consider the composition map

$$\mu :{\mathrm{Hom}}_{\mathcal{A}}(X,Y)\otimes {\mathrm{Hom}}_{\mathcal{A}}(Y,Z)⟶{\mathrm{Hom}}_{\mathcal{A}}(X,Z).$$

Here

$${\mathrm{Hom}}_{\mathcal{A}}(X,Y)\otimes {\mathrm{Hom}}_{\mathcal{A}}(Y,Z)\cong \underset{i,j,j^{\prime },s}{⨁}{\mathrm{Hom}}_{\mathcal{A}}(X_i,Y_j)\otimes {\mathrm{Hom}}_{\mathcal{A}}(Y_{j^{\prime }},Z_s),$$

and

$${\mathrm{Hom}}_{\mathcal{A}}(X,Z)\cong \underset{i,s}{⨁}{\mathrm{Hom}}_{\mathcal{A}}(X_i,Z_s).$$

However, $\mu$$\mu$ vanishes on mismatched indices, so it effectively factors through

$$\underset{i,j,s}{⨁}{\mathrm{Hom}}_{\mathcal{A}}(X_i,Y_j)\otimes {\mathrm{Hom}}_{\mathcal{A}}(Y_j,Z_s)⟶\underset{i,s}{⨁}{\mathrm{Hom}}_{\mathcal{A}}(X_i,Z_s).$$

On each direct summand, ${\mu }_{i,j,s}(f_{j,i}\otimes g_{s,j})=g_{s,j}\circ f_{j,i}$$\mu_{i,j,s}(f_{j,i} \otimes g_{s,j}) = g_{s,j} \circ f_{j,i}$. Summing up yields

$$\sum _{s,i}\sum _j{\iota }_s\circ g_{s,j}\circ f_{j,i}\circ {\pi }_i,$$

which is precisely matrix multiplication.