A Homological Approach to Harmonic Decomposition: From Yoneda to Poincaré

Harmonic functions via modules and $\operatorname{Ext}^1$A short exact sequence of quotientsThe long exact sequenceComputing ${\mathrm{Ext}}_R^1(\mathbb{C},M)$CorollaryCollaboration Report: With DeepSeek V41. Overview2. Contributions of the Author3. Contributions of the Assistant4. Character of the Collaboration5. Closing Remarks

 

Harmonic functions via modules and ${\mathrm{Ext}}^1$$\operatorname{Ext}^1$

Let $R$$R$ be a commutative ring and $\mathsf{Mod}(R)$$\mathsf{Mod}(R)$ the category of $R$$R$-modules. The identity functor is represented by $R$$R$ itself:

$${\mathsf{id}}_R\cong {\mathrm{Hom}}_R(R,-).$$

Yoneda’s Lemma gives $\mathrm{Nat}({\mathrm{Hom}}_R(R,-),{\mathrm{Hom}}_R(R,-))\cong R$$\operatorname{Nat}(\operatorname{Hom}_R(R,-),\,\operatorname{Hom}_R(R,-)) \cong R$.

For any $a\in R$$a \in R$, the kernel of multiplication by $a$$a$ on the represented functor is again representable:

$$\ker ({\mathrm{Hom}}_R(R,-)\stackrel{a\cdot -}{\to }{\mathrm{Hom}}_R(R,-))\cong {\mathrm{Hom}}_R(R/(a),-).$$

We specialise to $R=\mathbb{C}[X,Y]$$R = \mathbb C[X,Y]$ and $a=X^2+Y^2=(X+iY)(X-iY)$$a = X^2+Y^2 = (X+iY)(X-iY)$. Turn $C^{\mathrm{\infty }}({\mathbb{R}}^2,\mathbb{C})$$C^\infty(\mathbb R^2,\mathbb C)$ into an $R$$R$-module by letting polynomials act as differential operators:

$$X\cdot f={\partial }_{x}f,Y\cdot f={\partial }_{y}f$$

(identify ${\mathbb{R}}^2\cong \mathbb{C}$$\mathbb R^2 \cong \mathbb C$ so that multiplication by $i$$i$ is rotation by ${90}^{\circ }$$90^\circ$). Then

$$\mathcal{H}({\mathbb{R}}^2):=\text{harmonic functions on }{\mathbb{R}}^2\cong {\mathrm{Hom}}_R(R/(X^2+Y^2),C^{\mathrm{\infty }}({\mathbb{R}}^2,\mathbb{C})).$$

Now look at the map

$$R⟶R/(X+iY)\oplus R/(X-iY),f\mapsto (fmod(X+iY),fmod(X-iY)).$$

Its kernel is exactly $(X^2+Y^2)$$(X^2+Y^2)$, so we get an injection

$$i:R/(X^2+Y^2)\hookrightarrow R/(X+iY)\oplus R/(X-iY),1\mapsto (1,1).$$

Apply the contravariant functor ${\mathrm{Hom}}_R(-,C^{\mathrm{\infty }}({\mathbb{R}}^2,\mathbb{C}))$$\operatorname{Hom}_R\bigl(-,\,C^\infty(\mathbb R^2,\mathbb C)\bigr)$ to obtain

$$i^{\ast }:{\mathrm{Hom}}_R(R/(X-iY),C^{\mathrm{\infty }})\oplus {\mathrm{Hom}}_R(R/(X+iY),C^{\mathrm{\infty }})⟶\mathcal{H}({\mathbb{R}}^2),$$

$(u,v)\mapsto u+v$$(u,v) \mapsto u+v$.

Set

$$\mathcal{O}({\mathbb{R}}^2)={\mathrm{Hom}}_R(R/(X-iY),C^{\mathrm{\infty }}),\bar{\mathcal{O}}({\mathbb{R}}^2)={\mathrm{Hom}}_R(R/(X+iY),C^{\mathrm{\infty }}),$$

the spaces of entire holomorphic and anti‑holomorphic functions on $\mathbb{C}\cong {\mathbb{R}}^2$$\mathbb C \cong \mathbb R^2$. The kernel of $i^{\ast }$$i^*$ consists of pairs $(u,v)$$(u,v)$ with $u+v=0$$u+v=0$ and $({\partial }_x-i{\partial }_y)u=0$$(\partial_x-i\partial_y)u=0$, $({\partial }_x+i{\partial }_y)v=0$$(\partial_x+i\partial_y)v=0$. It follows that $u$$u$ is both holomorphic and anti‑holomorphic, hence constant, and $v=-u$$v=-u$. Thus $\ker i^{\ast }\cong \mathbb{C}$$\ker i^* \cong \mathbb C$ via $c\mapsto (c,-c)$$c \mapsto (c,-c)$.

So far we have an exact sequence

$$0⟶\mathbb{C}⟶\mathcal{O}({\mathbb{R}}^2)\oplus \bar{\mathcal{O}}({\mathbb{R}}^2)\stackrel{+}{\to }\mathcal{H}({\mathbb{R}}^2).$$

To see that the addition map is surjective we study the cokernel with ${\mathrm{Ext}}^1$$\operatorname{Ext}^1$.

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A short exact sequence of quotients

Put $I=(X+iY)$$I = (X+iY)$, $J=(X-iY)$$J = (X-iY)$. The two polynomials are coprime, so $I\cap J=IJ=(X^2+Y^2)$$I \cap J = IJ = (X^2+Y^2)$, while $I+J=(X,Y)$$I+J = (X,Y)$. For any two ideals there is a canonical short exact sequence

$$0⟶R/(I\cap J)\stackrel{f}{⟶}R/I\times R/J\stackrel{g}{⟶}R/(I+J)⟶0,$$

where $f(r+I\cap J)=(r+I,r+J)$$f(r+I\cap J) = (r+I,\, r+J)$ and $g(a+I,b+J)=(a-b)+(I+J)$$g(a+I,\, b+J) = (a-b)+(I+J)$. The map $g$$g$ is visibly onto, and its kernel is exactly the image of $f$$f$. (This is the general version of the Chinese Remainder Theorem when the ideals are not coprime.)

In our case this becomes

$$0\to R/(X^2+Y^2)\stackrel{i}{\to }R/(X+iY)\times R/(X-iY)\stackrel{p}{\to }\mathbb{C}\to 0,$$

where $\mathbb{C}\cong R/(X,Y)$$\mathbb C \cong R/(X,Y)$ and $p$$p$ is “take the difference of the constant terms”.

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The long exact sequence

Let $M=C^{\mathrm{\infty }}({\mathbb{R}}^2,\mathbb{C})$$M = C^\infty(\mathbb R^2,\mathbb C)$. Applying ${\mathrm{Hom}}_R(-,M)$$\operatorname{Hom}_R(-,M)$ to the short exact sequence above gives a long exact sequence containing

$$0\to {\mathrm{Hom}}_R(\mathbb{C},M)⟶\mathcal{O}({\mathbb{R}}^2)\oplus \bar{\mathcal{O}}({\mathbb{R}}^2)\stackrel{+}{\to }\mathcal{H}({\mathbb{R}}^2)⟶{\mathrm{Ext}}_R^1(\mathbb{C},M)⟶\cdots$$

Now ${\mathrm{Hom}}_R(\mathbb{C},M)\cong \{f\in M\mid (X,Y)f=0\}$$\operatorname{Hom}_R(\mathbb C,M) \cong \{f \in M \mid (X,Y)f=0\}$, the constant functions, so it is a copy of $\mathbb{C}$$\mathbb C$ mapped by $c\mapsto (c,-c)$$c\mapsto (c,-c)$ as before.

The addition map is surjective if and only if the obstruction space ${\mathrm{Ext}}_R^1(\mathbb{C},M)$$\operatorname{Ext}^1_R(\mathbb C,M)$ vanishes.

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Computing ${\mathrm{Ext}}_R^1(\mathbb{C},M)$$\operatorname{Ext}^1_R(\mathbb C,M)$

The module $\mathbb{C}\cong R/(X,Y)$$\mathbb C \cong R/(X,Y)$ has a Koszul free resolution

$$\begin{array}{c}0⟶R\stackrel{\left(\begin{array}{c}Y\\ -X\end{array}\right)}{\to }{R}^2\stackrel{(X,Y)}{\to }R⟶\mathbb{C}⟶0.\end{array}$$

Delete the term $\mathbb{C}$$\mathbb C$ and apply ${\mathrm{Hom}}_R(-,M)$$\operatorname{Hom}_R(-,M)$; using ${\mathrm{Hom}}_R(R,M)\cong M$$\operatorname{Hom}_R(R,M) \cong M$ and ${\mathrm{Hom}}_R(R^2,M)\cong M^2$$\operatorname{Hom}_R(R^2,M) \cong M^2$ we obtain the cochain complex

$$0⟶M\stackrel{({\partial }_x,{\partial }_y)}{\to }{M}^2\stackrel{(\xi ,\eta )\mapsto {\partial }_y\xi -{\partial }_x\eta }{\to }M⟶0.$$

By definition,

$${\mathrm{Ext}}_R^1(\mathbb{C},M)=\frac{\ker (M^2\to M)}{\mathrm{im}(M\to M^2)}=\frac{\{(\xi ,\eta )\in M^2\mid {\partial }_y\xi -{\partial }_x\eta =0\}}{\{({\partial }_{x}f,{\partial }_{y}f)\mid f\in M\}}.$$

To see what this quotient is, identify $(\xi ,\eta )$$(\xi,\eta)$ with the smooth $1$$1$-form $\omega =\xi dx+\eta dy$$\omega = \xi\,dx + \eta\,dy$ on ${\mathbb{R}}^2$$\mathbb R^2$. Its exterior derivative is $d\omega =({\partial }_x\eta -{\partial }_y\xi )dx\wedge dy$$d\omega = (\partial_x\eta - \partial_y\xi)\,dx\wedge dy$, so the condition

$${\partial }_y\xi -{\partial }_x\eta =0$$

means exactly $d\omega =0$$d\omega = 0$ — $\omega$$\omega$ is closed. The image of the first differential consists of the pairs $({\partial }_{x}f,{\partial }_{y}f)$$(\partial_x f, \partial_y f)$, which correspond to the exact $1$$1$-form $df$$df$. Hence the quotient is precisely the first de Rham cohomology of ${\mathbb{R}}^2$$\mathbb R^2$:

$${\mathrm{Ext}}_R^1(\mathbb{C},M)\cong H_{\mathrm{dR}}^1({\mathbb{R}}^2).$$

Since ${\mathbb{R}}^2$$\mathbb R^2$ is contractible, the Poincaré lemma tells us every closed $1$$1$-form is exact, so $H_{\mathrm{dR}}^1({\mathbb{R}}^2)=0$$H^1_{\mathrm{dR}}(\mathbb R^2)=0$ and therefore

$${\mathrm{Ext}}_R^1(\mathbb{C},M)=0.$$

Remark. For any simply connected domain $U\subseteq {\mathbb{R}}^2$$U \subseteq \mathbb R^2$, the first de Rham cohomology vanishes: $H_{\mathrm{dR}}^1(U)=0$$H^1_{\mathrm{dR}}(U)=0$. Hence the same argument yields the short exact sequence

$$0⟶\mathbb{C}⟶\mathcal{O}(U)\oplus \bar{\mathcal{O}}(U)\stackrel{+}{\to }\mathcal{H}(U)⟶0.$$

In particular, $\mathcal{O}(U)\subset \mathcal{H}(U)$$\mathcal O(U) \subset \mathcal H(U)$, and $\mathcal{H}(U)\cong (\mathcal{O}(U)\oplus \bar{\mathcal{O}}(U))/\mathbb{C}$$\mathcal H(U) \cong \bigl(\mathcal O(U) \oplus \overline{\mathcal O}(U)\bigr)/\mathbb C$, where $\mathbb{C}$$\mathbb C$ sits in the product as the constants $(c,-c)$$(c,-c)$.

Corollary

If $U\subseteq \mathbb{C}$$U \subseteq \mathbb C$ is simply connected, every harmonic function $h:U\to \mathbb{C}$$h \colon U \to \mathbb C$ splits as

$$h=f+\bar{g}$$

with $f,g$$f, g$ holomorphic on $U$$U$. The decomposition is unique up to an additive constant.

This follows from $+$$+$ is surjetive directly.

Construction. Write $\partial =\frac{1}{2}({\partial }_x-i{\partial }_y)$$\partial = \frac{1}{2}(\partial_x - i\partial_y)$ and $\overline{\partial }=\frac{1}{2}({\partial }_x+i{\partial }_y)$$\bar\partial = \frac{1}{2}(\partial_x + i\partial_y)$. The Laplacian factors as $\mathrm{\Delta }=4\overline{\partial }\partial$$\Delta = 4\bar\partial\partial$, so $h$$h$ harmonic means

$$\overline{\partial }(\partial h)=0.$$

This says that $\partial h$$\partial h$ is holomorphic. On a simply connected domain every holomorphic function has a primitive, so we can pick an $f\in \mathcal{O}(U)$$f \in \mathcal O(U)$ with $\partial f=\partial h$$\partial f = \partial h$ (for instance $\begin{array}{c}f(z)={\int }_{z_0}^z\partial h(\zeta )d\zeta \end{array}$$f(z) = \int_{z_0}^z \partial h(\zeta)\,d\zeta\\$, which is path‑independent.) Now set $\bar{g}=h-f$$\overline{g} = h - f$. $◻$$\square$

Collaboration Report: With DeepSeek V4

We recount how the author and the assistant worked together to produce the note A Homological Approach to Harmonic Decomposition: From Yoneda to Poincaré, highlighting the contributions of each side.

1. Overview

The project was driven by a clear vision: recast the classical decomposition of harmonic functions into holomorphic and anti‑holomorphic parts as a consequence of Yoneda’s lemma and a short exact sequence of modules over $R=\mathbb{C}[X,Y]$$R=\mathbb C[X,Y]$. The author provided the global strategy and the key algebraic insights; the assistant supplied technical verifications, homological computations, and stylistic polishing. The interaction was iterative – the author posed directional questions, the assistant filled in details, and together the final manuscript took shape.

2. Contributions of the Author

• Original idea and framework The author observed that the identity functor on $\mathsf{Mod}(R)$$\mathsf{Mod}(R)$ is represented by $R$$R$, and that for $a=X^2+Y^2$$a = X^2+Y^2$ the kernel of multiplication by $a$$a$ corresponds to ${\mathrm{Hom}}_R(R/(a),-)$$\operatorname{Hom}_R(R/(a),-)$. Interpreting $C^{\mathrm{\infty }}({\mathbb{R}}^2,\mathbb{C})$$C^\infty(\mathbb R^2,\mathbb C)$ as an $R$$R$-module via differential operators immediately turned the space of harmonic functions into ${\mathrm{Hom}}_R(R/(X^2+Y^2),C^{\mathrm{\infty }})$$\operatorname{Hom}_R(R/(X^2+Y^2), C^\infty)$. This algebraic starting point was entirely the author’s.

• Construction of the key maps The author introduced the injection

  $$i:R/(X^2+Y^2)\hookrightarrow R/(X+iY)\times R/(X-iY),1\mapsto (1,1),$$

  and recognised that applying ${\mathrm{Hom}}_R(-,C^{\mathrm{\infty }})$$\operatorname{Hom}_R(-,C^\infty)$ yields the addition map $i^{\ast }(u,v)=u+v$$i^*(u,v) = u+v$. The problem of decomposing harmonic functions was thus transformed into studying the kernel and cokernel of $i^{\ast }$$i^*$.

• Main conclusions and generalisations The author explicitly requested an ${\mathrm{Ext}}^1$$\operatorname{Ext}^1$ explanation of surjectivity, identified the obstruction with de Rham cohomology, and formulated the short exact sequence

  $$0\to \mathbb{C}\to \mathcal{O}\oplus \bar{\mathcal{O}}\to \mathcal{H}\to 0.$$

  The author further extended the result to arbitrary simply connected domains and wrote an independent operator‑factorisation proof.

• Final editing and style The author supplied a complete Markdown draft, repeatedly adjusted notation for consistency, insisted on $$...$$ for displayed equations, and refined the language to sound natural rather than formulaic. The final version owes its readability to the author’s persistent polishing.

3. Contributions of the Assistant

• Homological details The assistant explained the general short exact sequence

  $$0\to R/(I\cap J)\to R/I\times R/J\to R/(I+J)\to 0,$$

  verified that $I+J=(X,Y)$$I+J=(X,Y)$ yields the cokernel $\mathbb{C}$$\mathbb C$, and clarified its relation to the Chinese Remainder Theorem.

• Explicit computation of ${\mathrm{Ext}}^1$$\operatorname{Ext}^1$ The assistant provided a Koszul free resolution of $\mathbb{C}\cong R/(X,Y)$$\mathbb C \cong R/(X,Y)$, translated the $\mathrm{Hom}$$\operatorname{Hom}$-complex into the differential operator complex

  $$0\to M\stackrel{({\partial }_x,{\partial }_y)}{\to }{M}^2\stackrel{(\xi ,\eta )\mapsto {\partial }_y\xi -{\partial }_x\eta }{\to }M\to 0,$$

  and showed that its first cohomology is precisely closed $1$$1$-forms modulo exact $1$$1$-forms, i.e. $H_{\mathrm{dR}}^1({\mathbb{R}}^2)$$H^1_{\mathrm{dR}}(\mathbb R^2)$. The vanishing of this cohomology for a contractible domain completed the proof.

• Clarifications and corrections When the author raised uncertainties – “why can we drop the term $\mathbb{C}$$\mathbb C$?”, “what does the notation ${\partial }_y\cdot (1)-{\partial }_x\cdot (2)$$\partial_y\cdot(1)-\partial_x\cdot(2)$ mean?”, “is $g$$g$ a ring homomorphism?” – the assistant traced each issue back to definitions and suggested clearer alternatives. The distinction between deleting the resolved module $N$$N$ and deleting $\mathrm{Hom}(N,M)$$\operatorname{Hom}(N,M)$ was explicitly addressed, preventing a misunderstanding that $\mathrm{Hom}(\mathbb{C},M)=0$$\operatorname{Hom}(\mathbb C,M)=0$.

• Writing and structural advice Several English drafts were offered, progressively adjusted according to the author’s feedback: removing “AI‑flavour”, using $$ for displayed equations, breaking lines for readability, and choosing a concise title.

• Deeper commentary When the author remarked that operator factorisation feels more natural than the traditional approach, the assistant elaborated on how $\mathrm{\Delta }=4\overline{\partial }\partial$$\Delta = 4\bar\partial\partial$ provides a structural explanation complementary to the exact sequence – one gives existence, the other gives an explicit formula.

4. Character of the Collaboration

• Question‑driven The author continuously pushed with questions like “What is the deep reason?”, “Why is this quotient exactly de Rham?”, “How is this traditionally stated?”. These shaped the depth of the exposition.

• Iterative refinement Notation, language, and layout went through multiple rounds. The author maintained a strong vision for the final form; the assistant rapidly adapted technical passages to match that vision.

• Complementary strengths The author contributed the overarching narrative and algebraic‑geometric intuition; the assistant contributed the routine verifications of commutative and homological algebra. Together they built a complete bridge from Yoneda’s lemma to the Poincaré lemma.

5. Closing Remarks

The resulting blog post is a product of genuine collaboration: the author drew the map, the assistant made sure every path was solidly paved. The functorial viewpoint on harmonic decomposition is the author’s insight; its rigorous execution into a self‑contained piece of mathematics is shared work.