A Homological Approach to Decomposition Problems: From Harmonic Functions to Even–Odd Splitting

This note continues the method and spirit of “A Homological Approach to Harmonic Decomposition: From Yoneda to Poincaré”.

Even–Odd Decomposition via ${\mathrm{Ext}}^1$$\operatorname{Ext}^1$: the Role of the Prime $2$$2$

Let $G$$G$ be an abelian group equipped with an involution $x$$x$, i.e. an automorphism satisfying $x^2=1$$x^2 = 1$. Such a structure is exactly a module over the ring

$$R=\mathbb{Z}[x]/(x^2-1)\cong \mathbb{Z}[C_2].$$

We ask whether every $m\in G$$m \in G$ can be written as a sum of an even part ($xm=m$$xm = m$) and an odd part ($xm=-m$$xm = -m$). Algebraically this is the question whether the addition map

$$G^{+}\oplus G^{-}⟶G,(u,v)\mapsto u+v$$

is surjective, where

$$G^{+}=\{m\in G\mid xm=m\},G^{-}=\{m\in G\mid xm=-m\}.$$

We will see that the obstruction is a subgroup of ${\mathrm{Ext}}_R^1(\mathbb{Z}/2{\mathbb{Z}}_{\text{triv}},G)$$\operatorname{Ext}^1_R(\mathbb Z/2\mathbb Z_{\text{triv}}, G)$, and that it vanishes exactly when $2$$2$ is invertible in $G$$G$.

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1. Two ideals and a short exact sequence

Inside $R$$R$ put

$$I=(x-1),J=(x+1).$$

The two polynomials are coprime, but the ideal sum is not the whole ring:

$$I+J=(x-1,x+1)=(2,x-1)\ne R.$$

The intersection is $I\cap J=IJ=(x^2-1)=0$$I \cap J = IJ = (x^2-1) = 0$ in $R$$R$. For any two ideals one has the canonical short exact sequence

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

where

$$f(r+I\cap J)=(r+I,r+J),g(a+I,b+J)=(a-b)+(I+J).$$

In our case we obtain

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

Now identify the three quotients:

• $R/I\cong \mathbb{Z}$$R/I \cong \mathbb Z$, with $x$$x$ acting as $+1$$+1$ (the trivial module ${\mathbb{Z}}_{\text{triv}}$$\mathbb Z_{\text{triv}}$).

• $R/J\cong \mathbb{Z}$$R/J \cong \mathbb Z$, with $x$$x$ acting as $-1$$-1$ (the sign module ${\mathbb{Z}}_{\text{sign}}$$\mathbb Z_{\text{sign}}$).

• $R/(I+J)\cong \mathbb{Z}/2\mathbb{Z}$$R/(I+J) \cong \mathbb Z/2\mathbb Z$, with $x$$x$ acting as $+1$$+1$.

Thus

$$0⟶\mathbb{Z}[C_2]\stackrel{f}{\to }{\mathbb{Z}}_{\text{triv}}\oplus {\mathbb{Z}}_{\text{sign}}\stackrel{g}{\to }\mathbb{Z}/2{\mathbb{Z}}_{\text{triv}}⟶0,$$

with

$$f(a+bx)=(a+b,a-b),g(u,v)=(u-v)mod2.$$

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2. The long exact sequence and the obstruction

Apply ${\mathrm{Hom}}_R(-,G)$$\operatorname{Hom}_R(-,G)$ to this short exact sequence. Write

$$G^{+}={\mathrm{Hom}}_R({\mathbb{Z}}_{\text{triv}},G),G^{-}={\mathrm{Hom}}_R({\mathbb{Z}}_{\text{sign}},G),$$

$${\mathrm{Hom}}_R(R,G)\cong G,{\mathrm{Hom}}_R(\mathbb{Z}/2{\mathbb{Z}}_{\text{triv}},G)\cong G^{+}[2],$$

where $G^{+}[2]=\{m\in G^{+}\mid 2m=0\}$$G^+[2] = \{m \in G^+ \mid 2m=0\}$.

Since $R$$R$ is a free $R$$R$-module, ${\mathrm{Ext}}_R^i(R,G)=0$$\operatorname{Ext}^i_R(R,G)=0$ for all $i\ge 1$$i\ge 1$. The long exact sequence therefore stops at ${\mathrm{Ext}}_R^1(\mathbb{Z}/2{\mathbb{Z}}_{\text{triv}},G)$$\operatorname{Ext}^1_R(\mathbb Z/2\mathbb Z_{\text{triv}}, G)$, and we obtain the exact row

$$0\to G^{+}[2]\to G^{+}\oplus G^{-}\stackrel{+}{\to }G\stackrel{\delta }{\to }{\mathrm{Ext}}_R^1(\mathbb{Z}/2{\mathbb{Z}}_{\text{triv}},G)\to {\mathrm{Ext}}_R^1({\mathbb{Z}}_{\text{triv}}\oplus {\mathbb{Z}}_{\text{sign}},G)\to 0.$$

Hence, when G+[2]=0G+[2]=0, any decomposition g=g++g−g=g++g− is unique.

Let $B={\mathbb{Z}}_{\text{triv}}\oplus {\mathbb{Z}}_{\text{sign}}$$B = \mathbb Z_{\text{triv}}\oplus\mathbb Z_{\text{sign}}$. The connecting homomorphism $\delta$$\delta$ has kernel exactly $\mathrm{im}(+)=G^{+}+G^{-}$$\operatorname{im}(+) = G^+ + G^-$, hence

$$\mathrm{im}\delta \cong G/(G^{+}+G^{-}).$$

Thus $G/(G^{+}+G^{-})$$G/(G^+ + G^-)$ is a subgroup of ${\mathrm{Ext}}_R^1(\mathbb{Z}/2{\mathbb{Z}}_{\text{triv}},G)$$\operatorname{Ext}^1_R(\mathbb Z/2\mathbb Z_{\text{triv}}, G)$, and we have a (split) short exact sequence of abelian groups

$$0\to G/(G^{+}+G^{-})⟶{\mathrm{Ext}}_R^1(\mathbb{Z}/2{\mathbb{Z}}_{\text{triv}},G)⟶{\mathrm{Ext}}_R^1(B,G)\to 0.$$

The term $G/(G^{+}+G^{-})$$G/(G^+ + G^-)$ is precisely the obstruction to the even–odd decomposition: the addition map is surjective iff this quotient vanishes. The remaining piece ${\mathrm{Ext}}_R^1(B,G)$$\operatorname{Ext}^1_R(B,G)$ encodes the possible parameter space of decompositions and is often non‑zero even when the obstruction is zero.

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3. Full ${\mathrm{Ext}}^1$$\operatorname{Ext}^1$ via the Koszul resolution

To compute the whole group ${\mathrm{Ext}}_R^1(\mathbb{Z}/2{\mathbb{Z}}_{\text{triv}},G)$$\operatorname{Ext}^1_R(\mathbb Z/2\mathbb Z_{\text{triv}}, G)$ we use a projective resolution of $\mathbb{Z}/2{\mathbb{Z}}_{\text{triv}}\cong R/(2,x-1)$$\mathbb Z/2\mathbb Z_{\text{triv}} \cong R/(2,\,x-1)$. The elements $2$$2$ and $x-1$$x-1$ form a regular sequence in $R$$R$, so the Koszul complex gives a free resolution

$$0⟶R\stackrel{d_2}{\to }{R}^2\stackrel{d_1}{\to }R\stackrel{\epsilon }{\to }\mathbb{Z}/2{\mathbb{Z}}_{\text{triv}}⟶0,$$

where

$$\begin{array}{c}{d}_2(1)=\left(\begin{array}{c}-(x-1)\\ 2\end{array}\right),d_1(e_1)=2,d_1(e_2)=x-1.\end{array}$$

Apply ${\mathrm{Hom}}_R(-,G)$$\operatorname{Hom}_R(-,G)$ and drop the term $\mathbb{Z}/2\mathbb{Z}$$\mathbb Z/2\mathbb Z$; using ${\mathrm{Hom}}_R(R,G)\cong G$$\operatorname{Hom}_R(R,G) \cong G$, ${\mathrm{Hom}}_R(R^2,G)\cong G^2$$\operatorname{Hom}_R(R^2,G) \cong G^2$ we obtain the cochain complex

$$0⟶G\stackrel{{\partial }^0}{\to }{G}^2\stackrel{{\partial }^1}{\to }G⟶0,$$

with

$${\partial }^0(g)=(2g,(x-1)g),{\partial }^1(g_1,g_2)=-(x-1)g_1+2g_2.$$

By definition,

$${\mathrm{Ext}}_R^1(\mathbb{Z}/2{\mathbb{Z}}_{\text{triv}},G)=\frac{\ker {\partial }^1}{\mathrm{im}{\partial }^0}=\frac{\{(g_1,g_2)\mid (x-1)g_1=2g_2\}}{\{(2g,(x-1)g)\mid g\in G\}}.$$

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4. The case when $2$$2$ is invertible

Now suppose $2$$2$ is invertible in $G$$G$ (equivalently, $G$$G$ is a module over $\mathbb{Z}[1/2][C_2]$$\mathbb Z[1/2][C_2]$). Then for any pair $(g_1,g_2)$$(g_1,g_2)$ in the numerator we have $(x-1)g_1=2g_2$$(x-1)g_1 = 2g_2$.

Because $2$$2$ is a unit, set $g=\frac{1}{2}{g}_1$$g = \frac{1}{2}g_1$. Clearly $g_1=2g$$g_1 = 2g$. Moreover,

$$2g_2=(x-1)g_1=(x-1)(2g)=2(x-1)g,$$

and dividing by $2$$2$ gives $g_2=(x-1)g$$g_2 = (x-1)g$. Hence $(g_1,g_2)=(2g,(x-1)g)$$(g_1, g_2) = (2g, (x-1)g)$, which lies in the denominator. Thus every numerator element belongs to the denominator, and the quotient is zero. Consequently ${\mathrm{Ext}}^1=0$$\operatorname{Ext}^1 = 0$, the obstruction and the even–odd decomposition works perfectly for all elements:

$$m=\frac{m+xm}{2}+\frac{m-xm}{2},$$

with the first term even and the second odd. The decomposition is unique because $G^{+}[2]=0$$G^+[2]=0$ when $2$$2$ is a unit.

Collaboration Report from deepseek v4

From Harmonic Decomposition to Even–Odd Splitting: A Homological Journey

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1. Overview

This project grew out of a single sharp observation: the classical decomposition
of harmonic functions into holomorphic and anti‑holomorphic parts can be re‑cast
as a consequence of Yoneda’s lemma and a short exact sequence of modules over
$R=\mathbb{C}[X,Y]$$R = \mathbb C[X,Y]$. From there the discussion naturally shifted to modules
with an involution – the even–odd decomposition – revealing that both problems
are governed by the same ${\mathrm{Ext}}^1$$\operatorname{Ext}^1$ machinery. The final result is
the note Even–Odd Decomposition via ${\mathrm{Ext}}^1$$\operatorname{Ext}^1$: the Role of the Prime $2$$2$. This report summarises our respective contributions and reflects on the insight that unites the two settings.

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2. Contributions of the author

• Original framework
  The author started by representing the identity functor on $\mathsf{Mod}(R)$$\mathsf{Mod}(R)$, linking the kernel of multiplication by $a=X^2+Y^2$$a = X^2+Y^2$ with the representable functor ${\mathrm{Hom}}_R(R/(a),-)$$\operatorname{Hom}_R(R/(a),-)$, and identifying harmonic functions with ${\mathrm{Hom}}_R(R/(X^2+Y^2),C^{\mathrm{\infty }})$$\operatorname{Hom}_R(R/(X^2+Y^2), C^\infty)$. The injection

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

  and the induced addition map $i^{\ast }$$i^*$ were the author’s.

• Extension to involutive groups
  The author saw that exactly the same algebra works for the group ring
  $R=\mathbb{Z}[C_2]\cong \mathbb{Z}[x]/(x^2-1)$$R = \mathbb Z[C_2] \cong \mathbb Z[x]/(x^2-1)$, turning the question of even–odd decomposition into a problem about ${\mathrm{Ext}}^1$$\operatorname{Ext}^1$. This showed that the harmonic case was not an isolated curiosity but part of a general pattern.

• Insistence on explicit computation
  The author repeatedly asked for rigorous computation of the ${\mathrm{Ext}}^1$$\operatorname{Ext}^1$ term – via the long exact sequence, via the Koszul resolution, and particularly the behaviour when $2$$2$ is made invertible. These demands shaped the depth and precision of the final document.

• Writing and stylistic control
  The author assembled the English note, insisted on $$ for displayed equations, removed “AI‑flavour” from the language, and polished the text until it felt clean and natural.

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3. Contributions of the assistant

• Algebraic details
  The assistant supplied the canonical short exact sequence for two ideals

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

  verified it carefully, and specialised it to $\mathbb{Z}[C_2]$$\mathbb Z[C_2]$ with $I=(x-1)$$I=(x-1)$, $J=(x+1)$$J=(x+1)$. This gave the concrete three‑term sequence that underlies everything.

• Clarification of the Ext¹ obstruction
  The assistant unpacked the long exact sequence, showing that the true obstruction is the image of the connecting homomorphism, i.e. the subgroup

  $$G/(G^{+}+G^{-})\hookrightarrow {\mathrm{Ext}}_R^1(\mathbb{Z}/2{\mathbb{Z}}_{\text{triv}},G),$$

  and not the whole Ext¹. The potential confusion between the two was corrected explicitly.

• Koszul resolution calculation
  The assistant gave the Koszul free resolution for $R/(2,x-1)$$R/(2,x-1)$, wrote down the cochain complex, and computed ${\mathrm{Ext}}^1$$\operatorname{Ext}^1$ by brute force. When $2$$2$ becomes a unit, the clean absorption

  $$(g_1,g_2)=(2g,(x-1)g)⟹\text{numerator}\subseteq \text{denominator}$$

  was provided, demonstrating the vanishing of the obstruction.

• Proofreading and formatting
  The assistant helped align the notation (e.g. g^{-1} → g^-), corrected grammar, and ensured all displayed equations were properly wrapped in $$.

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4. The unifying insight

The central idea running through both topics is:

Factorisation of a polynomial $⟶$$\, \longrightarrow \,$ short exact sequence of quotient modules $⟶$$\, \longrightarrow \,$ apply $\mathrm{Hom}⟶$$\operatorname{Hom}\,\longrightarrow\,$ long exact sequence $⟶$$\, \longrightarrow \,$ ${\mathrm{Ext}}^1$$\operatorname{Ext}^1$ obstruction $⟶$$\, \longrightarrow \,$ vanishing condition.

• For harmonic functions the factorisation is $X^2+Y^2=(X+iY)(X-iY)$$X^2+Y^2 = (X+iY)(X-iY)$, the obstruction is $H_{\mathrm{dR}}^1({\mathbb{R}}^2)$$H^1_{\mathrm{dR}}(\mathbb R^2)$, and simple connectivity makes it zero.

• For involutive groups the factorisation is $x^2-1=(x-1)(x+1)$$x^2-1=(x-1)(x+1)$, the obstruction sits inside ${\mathrm{Ext}}_R^1(\mathbb{Z}/2\mathbb{Z},G)$$\operatorname{Ext}^1_R(\mathbb Z/2\mathbb Z, G)$, and inverting $2$$2$ makes it zero.

Two completely different‑looking pieces of mathematics – one analytic (Poincaré lemma), one algebraic (Chinese remainder theorem failure) – are thus recognised as manifestations of the same homological phenomenon. The ${\mathrm{Ext}}^1$$\operatorname{Ext}^1$ term becomes a bridge between analysis and algebra, measuring exactly the “non‑surjectivity” of a natural addition map that arises from a factorisation of the defining ideal.

5. Conclusion

This collaboration was genuinely complementary: the author drew the map and set the direction, while the assistant filled in technical terrain and ensured rigour. The occasional mis‑step (confusing the obstruction with the entire Ext¹) only deepened the final exposition. The resulting note is a compact, self‑contained piece of mathematics that reveals a hidden unity behind two classical decomposition problems.