[proofplan] We work in the model $T^2 = \mathbb{R}^2/(2\pi\mathbb{Z})^2$ with global coordinates $(x, y)$ inherited from $\mathbb{R}^2$. Smooth forms on $T^2$ correspond bijectively to smooth doubly $2\pi$-periodic forms on $\mathbb{R}^2$. Since $\dim T^2 = 2$, the spaces $\Omega^k(T^2)$ vanish for $k \geq 3$, so only degrees $0$, $1$, and $2$ are nontrivial. We compute $H^0$ from connectedness, $H^2$ from the integration pairing $\int_{T^2}$, and $H^1$ by [Fourier series](/page/Fourier%20Series) on $T^2$: the closed $1$-forms are those whose Fourier coefficients satisfy a single algebraic relation, and the exact $1$-forms are precisely those whose constant Fourier mode vanishes, so the quotient is parameterised by the pair of constant modes $(\hat a_{00}, \hat b_{00}) \in \mathbb{R}^2$. [/proofplan]

[step:Set up coordinates and identify smooth forms on $T^2$ with doubly periodic data] Let $\pi: \mathbb{R}^2 \to T^2 = \mathbb{R}^2 / (2\pi\mathbb{Z})^2$ denote the quotient map. Since $\pi$ is a smooth covering, pullback yields a bijection between $\Omega^k(T^2)$ and the space of $k$-forms on $\mathbb{R}^2$ invariant under translation by every element of $(2\pi\mathbb{Z})^2$. Writing $(x, y)$ for the standard coordinates on $\mathbb{R}^2$, and writing $\mathcal{L}^n$ for the $n$-dimensional Lebesgue measure on $\mathbb{R}^n$ (so that $\mathcal{L}^1$ and $\mathcal{L}^2$ denote Lebesgue measure in one and two variables, respectively), we identify \begin{align*} \Omega^0(T^2) &= C^\infty_{\mathrm{per}}(\mathbb{R}^2) := \{ f \in C^\infty(\mathbb{R}^2) : f(x + 2\pi, y) = f(x, y + 2\pi) = f(x, y)\}, \\ \Omega^1(T^2) &= \{ a\, dx + b\, dy : a, b \in C^\infty_{\mathrm{per}}(\mathbb{R}^2)\}, \\ \Omega^2(T^2) &= \{ f\, dx \wedge dy : f \in C^\infty_{\mathrm{per}}(\mathbb{R}^2)\}, \\ \Omega^k(T^2) &= 0 \quad \text{for } k \geq 3. \end{align*} The last line follows because $\dim_{\mathbb{R}} \Lambda^k T_p^* T^2 = \binom{2}{k} = 0$ for $k \geq 3$ at every point $p \in T^2$. The [exterior derivative](/theorems/1525) acts by $d(f) = \partial_x f\, dx + \partial_y f\, dy$ on $0$-forms and by $d(a\, dx + b\, dy) = (\partial_x b - \partial_y a)\, dx \wedge dy$ on $1$-forms. [/step]

[step:Conclude $H^k_{\mathrm{dR}}(T^2) = 0$ for $k \geq 3$ from dimension] For $k \geq 3$, $\Omega^k(T^2) = 0$ by the previous step, so both the cocycles $Z^k := \ker(d|_{\Omega^k})$ and coboundaries $B^k := d(\Omega^{k-1})$ vanish. Hence $H^k_{\mathrm{dR}}(T^2) = Z^k/B^k = 0$. [/step]

[step:Compute $H^0_{\mathrm{dR}}(T^2) \cong \mathbb{R}$ from connectedness] A closed $0$-form is a smooth function $f: T^2 \to \mathbb{R}$ with $df = 0$, i.e., $\partial_x f = 0$ and $\partial_y f = 0$ identically. Since $T^2$ is connected, this forces $f$ to be constant. Conversely, every constant function is closed (and there are no exact $0$-forms, as $\Omega^{-1} = 0$). Hence \begin{align*} H^0_{\mathrm{dR}}(T^2) = \{f \in C^\infty(T^2) : df = 0\} = \mathbb{R} \cdot \mathbf{1}, \end{align*} which is one-dimensional. The isomorphism $H^0_{\mathrm{dR}}(T^2) \to \mathbb{R}$ sends a constant function to its value. [/step]

[step:Set up Fourier series on $T^2$ and record the smoothness–decay correspondence]For each $f \in C^\infty(T^2)$ and each $(m, n) \in \mathbb{Z}^2$, define the Fourier coefficient \begin{align*} \widehat{f}: \mathbb{Z}^2 &\to \mathbb{C}, \\ (m, n) &\mapsto \frac{1}{4\pi^2} \int_{[0,2\pi]^2} f(x, y)\, e^{-i(mx + ny)} \, d\mathcal{L}^2(x, y). \end{align*} We record the following standard correspondence, which we justify in-line below: a sequence $(c_{mn})_{(m,n) \in \mathbb{Z}^2} \subset \mathbb{C}$ arises as the Fourier coefficients of some $f \in C^\infty(T^2)$ if and only if $(c_{mn})$ is *rapidly decreasing*, meaning that for every $N \geq 0$ there exists $C_N > 0$ with \begin{align*} |c_{mn}| \leq C_N (1 + m^2 + n^2)^{-N} \qquad \text{for all } (m, n) \in \mathbb{Z}^2, \end{align*} and in that case $f(x, y) = \sum_{(m,n) \in \mathbb{Z}^2} c_{mn}\, e^{i(mx + ny)}$ with absolute and [uniform convergence](/page/Uniform%20Convergence) of every partial-derivative series. The forward direction follows from repeated [integration by parts](/theorems/2098): for any $k, \ell \geq 0$, integrating by parts $k$ times in $x$ and $\ell$ times in $y$ (boundary terms vanish by $2\pi$-periodicity) gives $(im)^k (in)^\ell\, \widehat{f}(m, n) = \widehat{\partial_x^k \partial_y^\ell f}(m, n)$, and since $\partial_x^k \partial_y^\ell f$ is continuous on the compact torus, $|\widehat{\partial_x^k \partial_y^\ell f}(m,n)| \leq \|\partial_x^k \partial_y^\ell f\|_\infty$, yielding $|m|^k |n|^\ell |\widehat{f}(m,n)| \leq \|\partial_x^k \partial_y^\ell f\|_\infty$ for all $(m,n)$; choosing $k, \ell$ appropriately gives the rapid-decrease bound. The converse is obtained by applying the Weierstrass $M$-test to the formal derivative series, since each derivative only multiplies $c_{mn}$ by a polynomial in $m$ and $n$, which is dominated by the rapid-decrease bound. Real-valuedness of $f$ corresponds to $\overline{c_{mn}} = c_{-m,-n}$. Differentiation acts on Fourier coefficients by $\widehat{\partial_x f}(m,n) = im\, \widehat{f}(m,n)$ and $\widehat{\partial_y f}(m,n) = in\, \widehat{f}(m,n)$.[/step]

[guided]We will compute $H^1$ and $H^2$ by reducing the cohomology condition to an algebraic condition on Fourier coefficients, then inverting differentiation termwise. For this to be rigorous on smooth — as opposed to merely $L^2$ — forms, we need the precise dictionary between smoothness of $f$ and decay of $\widehat{f}$. For each $f \in C^\infty(T^2)$ and $(m, n) \in \mathbb{Z}^2$, define \begin{align*} \widehat{f}: \mathbb{Z}^2 &\to \mathbb{C}, \\ (m, n) &\mapsto \frac{1}{4\pi^2} \int_{[0,2\pi]^2} f(x, y)\, e^{-i(mx + ny)} \, d\mathcal{L}^2(x, y). \end{align*} Why is the map $f \mapsto \widehat{f}$ a bijection onto the rapidly decreasing sequences? Repeated [integration by parts](/theorems/210) in $x$ (using $2\pi$-periodicity to kill the boundary terms) gives \begin{align*} (im)^k\, \widehat{f}(m,n) = \widehat{\partial_x^k f}(m,n), \end{align*} and similarly for $y$. Since $\partial_x^k \partial_y^\ell f$ is continuous on the compact torus and thus bounded, $|\widehat{f}(m,n)|$ decays faster than any inverse polynomial in $|m| + |n|$, which is the rapid-decrease condition. Conversely, given a rapidly decreasing sequence $(c_{mn})$, the series $\sum c_{mn}\, e^{i(mx+ny)}$ and all its formal partial-derivative series converge absolutely and uniformly (a single application of Weierstrass $M$-test, with the rapid-decrease bound dominating any polynomial factor from differentiation), so the sum defines a smooth function $f$ with $\widehat{f} = (c_{mn})$. This is the version of the Schwartz-class duality $\mathcal{S}(\mathbb{Z}^n) \leftrightarrow C^\infty(T^n)$ that underpins the convergence statements in [Pointwise Convergence of Fourier Series](/theorems/1378) and [Fejér's Theorem](/theorems/584). The differentiation formulas $\widehat{\partial_x f} = im\, \widehat{f}$ and $\widehat{\partial_y f} = in\, \widehat{f}$ then follow from the same [integration by parts](/theorems/2098), and they translate exactness/closedness into linear algebra over $\mathbb{Z}^2$. In what follows, *all algebraic manipulations on Fourier coefficients are accompanied by a verification that the resulting sequence is rapidly decreasing*, which is what licenses the inverse [Fourier transform](/page/Fourier%20Transform) back to a smooth form.[/guided]

[step:Compute $H^2_{\mathrm{dR}}(T^2) \cong \mathbb{R}$ via integration]Define the integration map \begin{align*} I: \Omega^2(T^2) &\to \mathbb{R}, \\ f\, dx \wedge dy &\mapsto \int_{T^2} f\, dx \wedge dy = \int_{[0, 2\pi]^2} f(x, y)\, d\mathcal{L}^2(x, y). \end{align*} Since every $\omega \in \Omega^2(T^2)$ is closed (as $\Omega^3(T^2) = 0$), we have $Z^2 = \Omega^2(T^2)$, and we claim that $I$ induces an isomorphism $\bar{I}: H^2_{\mathrm{dR}}(T^2) \to \mathbb{R}$. **Vanishing on exact forms.** For $\eta = a\, dx + b\, dy \in \Omega^1(T^2)$, \begin{align*} I(d\eta) = \int_{[0,2\pi]^2} (\partial_x b - \partial_y a)\, d\mathcal{L}^2(x, y) = 0 \end{align*} by the [fundamental theorem of calculus](/theorems/632) applied separately in $x$ and $y$, using $2\pi$-periodicity of $a$ and $b$ to cancel boundary terms. Hence $I$ descends to $\bar{I}: H^2_{\mathrm{dR}}(T^2) \to \mathbb{R}$. **Surjectivity.** $\bar{I}([dx \wedge dy]) = 4\pi^2 \neq 0$. Since $\bar I$ is $\mathbb{R}$-linear and its codomain $\mathbb{R}$ is one-dimensional, having a nonzero value implies that $\bar I$ is surjective. **Injectivity.** Suppose $\omega = f\, dx \wedge dy \in Z^2$ satisfies $I(\omega) = 0$. Then $\widehat{f}(0, 0) = \frac{1}{4\pi^2} \int_{[0,2\pi]^2} f(x,y)\, d\mathcal{L}^2(x,y) = 0$. Define a $1$-form $\eta = a\, dx + b\, dy$ by specifying its Fourier coefficients: \begin{align*} \widehat{a}(m, n) &= \begin{cases} -\dfrac{\widehat{f}(0, n)}{in} & \text{if } m = 0 \text{ and } n \neq 0, \\ 0 & \text{otherwise,} \end{cases} \\ \widehat{b}(m, n) &= \begin{cases} \dfrac{\widehat{f}(m, n)}{im} & \text{if } m \neq 0, \\ 0 & \text{if } m = 0. \end{cases} \end{align*} Each of $(\widehat{a}(m,n))$ and $(\widehat{b}(m,n))$ is rapidly decreasing: indeed, $|\widehat{b}(m,n)| \leq |\widehat{f}(m,n)|$ for $|m| \geq 1$, and a rapidly decreasing sequence remains rapidly decreasing under such a uniform bound; similarly for $\widehat{a}$. These coefficients also satisfy the real-valuedness symmetry: for example, when $m \neq 0$, $\overline{\widehat b(m,n)} = \overline{\widehat f(m,n)/(im)} = \widehat f(-m,-n)/(i(-m)) = \widehat b(-m,-n)$, and the same computation with the defining cases gives $\overline{\widehat a(m,n)} = \widehat a(-m,-n)$. By the smoothness–decay correspondence from the previous step, there exist real-valued $a, b \in C^\infty(T^2)$ realising these coefficients. Then \begin{align*} \widehat{\partial_x b - \partial_y a}(m, n) = im\, \widehat{b}(m,n) - in\, \widehat{a}(m,n) = \begin{cases} \widehat{f}(m,n) & (m,n) \neq (0,0), \\ 0 & (m,n) = (0,0), \end{cases} \end{align*} which equals $\widehat{f}(m,n)$ for all $(m, n)$ since $\widehat{f}(0,0) = 0$ by hypothesis. By the uniqueness clause of the Fourier correspondence, $\partial_x b - \partial_y a = f$, hence $d\eta = \omega$. Therefore $\bar I$ is an isomorphism and $H^2_{\mathrm{dR}}(T^2) \cong \mathbb{R}$.[/step]

[guided]The strategy is to construct a *cohomological pairing* between $H^2_{\mathrm{dR}}(T^2)$ and $\mathbb{R}$ given by integration, and then to show it is an isomorphism. Two things must be checked: that integration descends to cohomology (i.e., kills exact forms), and that the descended map is bijective. Define the integration map \begin{align*} I: \Omega^2(T^2) &\to \mathbb{R}, \\ f\, dx \wedge dy &\mapsto \int_{T^2} f\, dx \wedge dy = \int_{[0, 2\pi]^2} f(x, y)\, d\mathcal{L}^2(x, y). \end{align*} Every $\omega \in \Omega^2(T^2)$ is closed automatically because $\Omega^3(T^2) = 0$, so $Z^2 = \Omega^2(T^2)$ and $H^2_{\mathrm{dR}}(T^2) = \Omega^2(T^2)/B^2$. **Why integration descends.** For $\eta = a\, dx + b\, dy \in \Omega^1(T^2)$, \begin{align*} I(d\eta) = \int_{[0,2\pi]^2} (\partial_x b - \partial_y a)\, d\mathcal{L}^2(x, y). \end{align*} We integrate $\partial_x b$ first in $x$: for fixed $y$, $\int_0^{2\pi} \partial_x b(x, y)\, d\mathcal{L}^1(x) = b(2\pi, y) - b(0, y) = 0$ by $2\pi$-periodicity in $x$. Likewise $\int_0^{2\pi} \partial_y a(x, y)\, d\mathcal{L}^1(y) = 0$ for fixed $x$. By Fubini, $I(d\eta) = 0$. This is the local version of [Stokes' theorem](/theorems/1530) on the boundaryless manifold $T^2$. Hence $I$ factors through $\bar I: H^2_{\mathrm{dR}}(T^2) \to \mathbb{R}$. **Surjectivity.** The class of $dx \wedge dy$ maps to $\int_{[0,2\pi]^2} 1\, d\mathcal{L}^2 = 4\pi^2 \neq 0$, so $\bar I$ hits a non-zero element of $\mathbb{R}$ and is therefore surjective (the codomain is one-dimensional). **Injectivity.** This is the substantive content: every closed $2$-form with zero total integral is exact. Suppose $\omega = f\, dx \wedge dy$ with $I(\omega) = 0$, i.e., $\widehat{f}(0, 0) = 0$ (since $\widehat{f}(0,0) = \frac{1}{4\pi^2}\int_{[0,2\pi]^2} f(x,y)\, d\mathcal{L}^2(x,y)$). We must construct $\eta = a\, dx + b\, dy$ with $d\eta = \omega$, i.e., $\partial_x b - \partial_y a = f$. In Fourier coefficients, this PDE becomes the algebraic equation \begin{align*} im\, \widehat{b}(m, n) - in\, \widehat{a}(m, n) = \widehat{f}(m, n) \qquad \text{for all } (m, n) \in \mathbb{Z}^2. \end{align*} For $(m, n) = (0, 0)$ both sides are zero — and this is *exactly* where the hypothesis $\widehat{f}(0,0) = 0$ enters. For $(m, n) \neq (0, 0)$ we have a single linear equation in two unknowns $(\widehat{a}(m,n), \widehat{b}(m,n))$, and we choose the simplest splitting: divide by $im$ when possible, otherwise by $-in$. This motivates the prescription \begin{align*} \widehat{a}(m, n) &= \begin{cases} -\dfrac{\widehat{f}(0, n)}{in} & m = 0,\ n \neq 0, \\ 0 & \text{otherwise,} \end{cases} \\ \widehat{b}(m, n) &= \begin{cases} \dfrac{\widehat{f}(m, n)}{im} & m \neq 0, \\ 0 & m = 0. \end{cases} \end{align*} We must check that this Fourier data corresponds to *smooth* $a, b$. For the $b$-data: when $|m| \geq 1$, \begin{align*} |\widehat{b}(m, n)| = \frac{|\widehat{f}(m,n)|}{|m|} \leq |\widehat{f}(m, n)|, \end{align*} and since $\widehat{f}$ is rapidly decreasing (because $f \in C^\infty(T^2)$), so is $\widehat{b}$. Similarly $|\widehat{a}(0, n)| = |\widehat{f}(0, n)|/|n| \leq |\widehat{f}(0, n)|$ for $|n| \geq 1$, so $\widehat{a}$ is rapidly decreasing. The coefficients satisfy the conjugate symmetry required for real-valued functions: $\overline{\widehat b(m,n)} = \widehat b(-m,-n)$ when $m \neq 0$ by the identity $\overline{1/(im)} = 1/(i(-m))$, while both sides vanish when $m = 0$; the verification for $\widehat a$ is identical on the line $m = 0$ and both sides vanish off that line. By the smoothness–decay correspondence, real-valued $a, b \in C^\infty(T^2)$ exist with these Fourier coefficients. We then compute, term by term in Fourier space, \begin{align*} \widehat{\partial_x b - \partial_y a}(m, n) &= im\, \widehat{b}(m,n) - in\, \widehat{a}(m,n) \\ &= \begin{cases} im \cdot \frac{\widehat{f}(m,n)}{im} = \widehat{f}(m,n) & m \neq 0, \\ -in \cdot \left( -\frac{\widehat{f}(0,n)}{in} \right) = \widehat{f}(0, n) & m = 0,\ n \neq 0, \\ 0 = \widehat{f}(0, 0) & (m, n) = (0, 0). \end{cases} \end{align*} In every case the value equals $\widehat{f}(m, n)$ (using $\widehat{f}(0, 0) = 0$ in the last case), so the Fourier coefficients of $\partial_x b - \partial_y a$ and $f$ agree. By uniqueness of [Fourier series](/page/Fourier%20Series) on $T^2$, $\partial_x b - \partial_y a = f$, hence $d\eta = \omega$. Combining surjectivity and injectivity, $\bar I: H^2_{\mathrm{dR}}(T^2) \to \mathbb{R}$ is an isomorphism.[/guided]

[step:Identify closed $1$-forms in Fourier coordinates] A $1$-form $\omega = a\, dx + b\, dy \in \Omega^1(T^2)$ has $d\omega = (\partial_x b - \partial_y a)\, dx \wedge dy$, so $\omega$ is closed if and only if $\partial_x b = \partial_y a$ identically on $T^2$. Translating into Fourier coefficients via $\widehat{\partial_x b}(m,n) = im\, \widehat{b}(m,n)$ and $\widehat{\partial_y a}(m,n) = in\, \widehat{a}(m,n)$, and using uniqueness of [Fourier series](/page/Fourier%20Series), $\omega$ is closed if and only if \begin{align*} m\, \widehat{b}(m, n) = n\, \widehat{a}(m, n) \qquad \text{for every } (m, n) \in \mathbb{Z}^2. \tag{$\ast$} \end{align*} Note that for $(m, n) = (0, 0)$ relation $(\ast)$ is automatic and imposes no constraint on $\widehat{a}(0, 0)$ and $\widehat{b}(0, 0)$. [/step]

[step:Identify exact $1$-forms in Fourier coordinates]A $1$-form $\omega = a\, dx + b\, dy$ is exact iff there exists $f \in C^\infty(T^2)$ with $df = \omega$, i.e., $\partial_x f = a$ and $\partial_y f = b$. In Fourier coefficients, this is the system \begin{align*} im\, \widehat{f}(m, n) = \widehat{a}(m, n), \qquad in\, \widehat{f}(m, n) = \widehat{b}(m, n) \qquad \text{for every } (m, n) \in \mathbb{Z}^2. \tag{$\ast\ast$} \end{align*} Evaluating $(\ast\ast)$ at $(m, n) = (0, 0)$ gives $\widehat{a}(0, 0) = 0$ and $\widehat{b}(0, 0) = 0$, which is therefore a *necessary* condition for exactness. We claim the converse: if $\omega$ is closed (so $(\ast)$ holds) and $\widehat{a}(0, 0) = \widehat{b}(0, 0) = 0$, then $\omega$ is exact. Define \begin{align*} \widehat{f}(m, n) := \begin{cases} \widehat{a}(m, n)/(im) & m \neq 0, \\ \widehat{b}(0, n)/(in) & m = 0,\ n \neq 0, \\ 0 & (m, n) = (0, 0). \end{cases} \end{align*} This sequence is rapidly decreasing: for $|m| \geq 1$, $|\widehat{f}(m,n)| \leq |\widehat{a}(m,n)|$, and for $m = 0, |n| \geq 1$, $|\widehat{f}(0,n)| \leq |\widehat{b}(0,n)|$, both rapidly decreasing because $a, b \in C^\infty(T^2)$. It also satisfies the real-valuedness symmetry $\overline{\widehat f(m,n)} = \widehat f(-m,-n)$: when $m \neq 0$ this follows from $\overline{\widehat a(m,n)/(im)} = \widehat a(-m,-n)/(i(-m))$, when $m = 0$ and $n \neq 0$ it follows from the corresponding symmetry for $\widehat b$, and at $(0,0)$ both sides equal $0$. By the smoothness–decay correspondence there exists a real-valued $f \in C^\infty(T^2)$ with these Fourier coefficients. We verify $(\ast\ast)$ on a case-by-case basis. For $m \neq 0$, $im\, \widehat{f}(m, n) = \widehat{a}(m, n)$ by definition, while $in\, \widehat{f}(m, n) = (in/im)\, \widehat{a}(m, n) = (n/m)\, \widehat{a}(m, n) = \widehat{b}(m, n)$ by the closedness relation $(\ast)$. For $m = 0$ and $n \neq 0$: $im\, \widehat{f}(0, n) = 0 = \widehat{a}(0, n)$, where the last equality uses $(\ast)$ at $(0, n)$ (which reads $0 = n\, \widehat{a}(0, n)$, so $\widehat{a}(0, n) = 0$ for $n \neq 0$); and $in\, \widehat{f}(0, n) = \widehat{b}(0, n)$ by definition. For $(m, n) = (0, 0)$ both equations read $0 = 0$ since $\widehat{a}(0, 0) = \widehat{b}(0, 0) = 0$. By uniqueness of [Fourier series](/page/Fourier%20Series), $\partial_x f = a$ and $\partial_y f = b$, so $df = \omega$, proving the converse.[/step]

[guided]We are setting up the linear algebra that will let us read off $H^1_{\mathrm{dR}}(T^2)$ directly. The closedness condition $\partial_x b = \partial_y a$ becomes a single equation per Fourier mode, while exactness imposes two equations per Fourier mode (one for each partial derivative of the potential $f$). The discrepancy lives entirely in the zero mode $(m, n) = (0, 0)$, and that is the source of the $\mathbb{R}^2$ in $H^1$. **Exactness in Fourier.** A $1$-form $\omega = a\, dx + b\, dy$ is exact iff there is $f \in C^\infty(T^2)$ with $df = \omega$, equivalently \begin{align*} \partial_x f = a, \qquad \partial_y f = b. \end{align*} Applying $\widehat{\,\cdot\,}$ and using $\widehat{\partial_x f}(m,n) = im\, \widehat{f}(m,n)$ and $\widehat{\partial_y f}(m,n) = in\, \widehat{f}(m,n)$, this becomes the system $(\ast\ast)$: \begin{align*} im\, \widehat{f}(m, n) = \widehat{a}(m, n), \qquad in\, \widehat{f}(m, n) = \widehat{b}(m, n) \qquad \text{for every } (m, n) \in \mathbb{Z}^2. \end{align*} At the zero mode $(m, n) = (0, 0)$, both equations read $0 = \widehat{a}(0,0)$ and $0 = \widehat{b}(0,0)$, so vanishing of the *constant* Fourier modes of $a$ and $b$ is *necessary* for exactness. Geometrically, $\widehat{a}(0, 0) = \frac{1}{4\pi^2}\int_{T^2} a\, d\mathcal{L}^2$, so we are saying that the *averages* of $a$ and $b$ are obstructions to exactness — which is consistent with the integrals $\int_{S^1 \times \{*\}} \omega$ and $\int_{\{*\} \times S^1} \omega$ along the two generating cycles being the natural cohomological pairing. **The converse: closed and zero-mean implies exact.** Suppose $\omega$ is closed (so $(\ast)$ holds) and $\widehat{a}(0, 0) = \widehat{b}(0, 0) = 0$. We construct a potential $f$ by directly solving the Fourier system. Off the zero mode we need to invert $im$ (or $in$ when $m = 0$); the closedness relation $(\ast)$ guarantees that the two formulas $\widehat{a}/(im)$ and $\widehat{b}/(in)$ agree when both are defined. We set \begin{align*} \widehat{f}(m, n) := \begin{cases} \widehat{a}(m, n)/(im) & m \neq 0, \\ \widehat{b}(0, n)/(in) & m = 0,\ n \neq 0, \\ 0 & (m, n) = (0, 0). \end{cases} \end{align*} *Smoothness check.* Why is $f$ smooth, and not merely $L^2$? For $|m| \geq 1$, $|\widehat{f}(m,n)| = |\widehat{a}(m,n)|/|m| \leq |\widehat{a}(m,n)|$, and for $m = 0$, $|n| \geq 1$, $|\widehat{f}(0,n)| = |\widehat{b}(0,n)|/|n| \leq |\widehat{b}(0,n)|$. The right-hand sides are rapidly decreasing because $a, b \in C^\infty(T^2)$, so $\widehat{f}$ is rapidly decreasing — which is the smoothness criterion from the previous step. We also check real-valuedness at the coefficient level. Since $a$ and $b$ are real-valued, $\overline{\widehat a(m,n)} = \widehat a(-m,-n)$ and $\overline{\widehat b(m,n)} = \widehat b(-m,-n)$. If $m \neq 0$, then $\overline{\widehat f(m,n)} = \overline{\widehat a(m,n)/(im)} = \widehat a(-m,-n)/(i(-m)) = \widehat f(-m,-n)$. If $m = 0$ and $n \neq 0$, the same computation with $\widehat b(0,n)/(in)$ gives $\overline{\widehat f(0,n)} = \widehat f(0,-n)$. At $(0,0)$ both sides equal $0$. Hence the inverse [Fourier series](/page/Fourier%20Series) defines a real-valued $f \in C^\infty(T^2)$. *Verification of $(\ast\ast)$.* - For $m \neq 0$: $im\, \widehat{f}(m,n) = im \cdot \widehat{a}(m,n)/(im) = \widehat{a}(m,n)$, which is the $\widehat{a}$-equation. For the $\widehat{b}$-equation, $in\, \widehat{f}(m,n) = (in/im)\, \widehat{a}(m,n) = (n/m)\, \widehat{a}(m,n)$; but $(\ast)$ reads $m\, \widehat{b}(m,n) = n\, \widehat{a}(m,n)$, so $(n/m)\, \widehat{a}(m,n) = \widehat{b}(m,n)$. Both equations are verified. - For $m = 0$, $n \neq 0$: the $\widehat{a}$-equation reads $0 = \widehat{a}(0, n)$. We verify this from $(\ast)$ at $(0, n)$: the relation reads $0 = n\, \widehat{a}(0, n)$, and since $n \neq 0$, $\widehat{a}(0, n) = 0$. The $\widehat{b}$-equation reads $in\, \widehat{f}(0, n) = \widehat{b}(0, n)$, which holds by definition of $\widehat{f}(0, n)$. - For $(m, n) = (0, 0)$: both equations read $0 = 0$, using $\widehat{a}(0,0) = \widehat{b}(0,0) = 0$. So $(\ast\ast)$ holds in every case. By uniqueness of Fourier expansions, $\partial_x f = a$ and $\partial_y f = b$, i.e., $df = \omega$. We have shown: $\omega \in B^1 \iff \omega \in Z^1$ and $\widehat{a}(0,0) = \widehat{b}(0,0) = 0$. This is precisely the algebraic input needed for the next step.[/guided]

[step:Conclude $H^1_{\mathrm{dR}}(T^2) \cong \mathbb{R}^2$] Define \begin{align*} J: Z^1(T^2) &\to \mathbb{R}^2, \\ \omega = a\, dx + b\, dy &\mapsto \bigl(\widehat{a}(0, 0),\, \widehat{b}(0, 0)\bigr) = \left( \frac{1}{4\pi^2}\int_{T^2} a\, d\mathcal{L}^2,\ \frac{1}{4\pi^2}\int_{T^2} b\, d\mathcal{L}^2 \right). \end{align*} (The image lies in $\mathbb{R}^2 \subset \mathbb{C}^2$ because $a, b$ are real-valued, hence $\widehat{a}(0,0), \widehat{b}(0,0) \in \mathbb{R}$.) $J$ is $\mathbb{R}$-linear. **Surjective.** $J(dx) = (1, 0)$ and $J(dy) = (0, 1)$, both closed. **Kernel.** By the analysis of the previous step, the closed form $\omega \in Z^1$ lies in $\ker J$ if and only if $\widehat{a}(0, 0) = \widehat{b}(0, 0) = 0$, which holds if and only if $\omega \in B^1$. Hence $\ker J = B^1$. By the [first isomorphism theorem](/theorems/791), $J$ induces an isomorphism \begin{align*} \bar J: H^1_{\mathrm{dR}}(T^2) = Z^1/B^1 \xrightarrow{\ \cong\ } \mathbb{R}^2. \end{align*} [/step]

[step:Assemble the four cohomology computations] Combining the previous steps: \begin{align*} H^0_{\mathrm{dR}}(T^2) &\cong \mathbb{R} && \text{(connectedness),} \\ H^1_{\mathrm{dR}}(T^2) &\cong \mathbb{R}^2 && \text{(via } \bar J\text{),} \\ H^2_{\mathrm{dR}}(T^2) &\cong \mathbb{R} && \text{(via } \bar I\text{),} \\ H^k_{\mathrm{dR}}(T^2) &= 0 \quad (k \geq 3) && \text{(dimension).} \end{align*} This is the claimed result. $\blacksquare$ [/step]