[proofplan] We compute the de Rham cohomology directly from the covering parametrisation $q:\mathbb{R}\to S^1$. Degree $0$ follows because closed smooth functions have zero derivative and $S^1$ is connected. Degrees $k\ge 2$ vanish because $S^1$ is one-dimensional. For degree $1$, we show that the angular form $\alpha$ is closed, not exact, and that every $1$-form differs from a unique constant multiple of $\alpha$ by an exact form; the constant is determined by the period around the circle. Throughout the proof, $\mathcal{L}^1$ denotes one-dimensional Lebesgue measure on $\mathbb{R}$. [/proofplan] [step:Identify closed functions on the connected circle with constants] Let $f\in \Omega^0(S^1)=C^\infty(S^1;\mathbb{R})$ satisfy $df=0$. We show that $f$ is constant. Define the smooth covering parametrisation \begin{align*} q:\mathbb{R}&\to S^1\\ t&\mapsto(\cos t,\sin t). \end{align*} The pullback $f\circ q:\mathbb{R}\to\mathbb{R}$ is smooth, and by the chain rule, \begin{align*} d(f\circ q)_t = q^*(df)_t = 0 \end{align*} for every $t\in \mathbb{R}$. Hence $f\circ q$ has derivative zero on the connected interval $\mathbb{R}$, so $f\circ q$ is constant. Since $q$ is surjective, $f$ is constant on $S^1$. Conversely, if $f:S^1\to\mathbb{R}$ is constant, then $df=0$. Therefore the closed $0$-forms are exactly the constant functions, and there are no exact $0$-forms except $0$. Thus \begin{align*} H^0_{\mathrm{dR}}(S^1)=\ker\left(d:\Omega^0(S^1)\to\Omega^1(S^1)\right)\cong \mathbb{R}. \end{align*} [guided] A de Rham $0$-cocycle is a smooth function whose differential vanishes. Let \begin{align*} f:S^1\to\mathbb{R} \end{align*} be a smooth function with $df=0$. To prove that $f$ is constant, it is convenient to pull it back to the real line using the standard parametrisation \begin{align*} q:\mathbb{R}&\to S^1\\ t&\mapsto(\cos t,\sin t). \end{align*} This map is smooth and surjective. The composite \begin{align*} f\circ q:\mathbb{R}\to\mathbb{R} \end{align*} is smooth. By the chain rule for pullbacks of differentials, \begin{align*} d(f\circ q)_t=q^*(df)_t=0 \end{align*} for every $t\in\mathbb{R}$. Thus the ordinary derivative of $f\circ q$ is zero everywhere on the connected set $\mathbb{R}$, so $f\circ q$ is constant. Since every point of $S^1$ is $q(t)$ for some $t\in\mathbb{R}$, the function $f$ itself is constant. The converse is immediate from the definition of the differential: a constant smooth function has zero differential. In the de Rham complex on $S^1$, there is no degree below $0$, so the image into $\Omega^0(S^1)$ is the zero subspace. Hence no nonzero $0$-form is exact, and \begin{align*} H^0_{\mathrm{dR}}(S^1)\cong \mathbb{R}, \end{align*} with the isomorphism sending a closed function to its constant value. [/guided] [/step] [step:Use one-dimensionality to eliminate all degrees at least two] Since $S^1$ is a smooth manifold of dimension $1$, the exterior power $\Lambda^kT_p^*S^1$ is the zero [vector space](/page/Vector%20Space) for every $p\in S^1$ and every integer $k\ge 2$. Therefore \begin{align*} \Omega^k(S^1)=0 \end{align*} for every $k\ge 2$. Hence \begin{align*} H^k_{\mathrm{dR}}(S^1)=0 \end{align*} for every $k\ge 2$. [guided] The reason all higher degrees vanish is purely dimensional. At each point $p\in S^1$, the tangent space $T_pS^1$ is a $1$-dimensional real vector space, so its cotangent space $T_p^*S^1$ is also $1$-dimensional. The alternating $k$-linear forms on a $1$-dimensional vector space vanish for every integer $k\ge 2$, because any $k$ vectors in a $1$-dimensional vector space are linearly dependent and an alternating form is zero on linearly dependent inputs. Thus \begin{align*} \Lambda^kT_p^*S^1=0 \end{align*} for every $p\in S^1$ and every integer $k\ge 2$. Therefore the vector space of smooth $k$-forms is \begin{align*} \Omega^k(S^1)=0. \end{align*} The de Rham cohomology group in degree $k\ge 2$ is a quotient of a subspace of $\Omega^k(S^1)$, so it is the zero vector space: \begin{align*} H^k_{\mathrm{dR}}(S^1)=0. \end{align*} [/guided] [/step] [step:Define the angular form and compute its pullback] Define the smooth $1$-form $\alpha\in\Omega^1(S^1)$ by \begin{align*} \alpha_p(v)=-p_2v_1+p_1v_2 \end{align*} for $p=(p_1,p_2)\in S^1$ and $v=(v_1,v_2)\in T_pS^1\subset\mathbb{R}^2$. For the parametrisation $q:\mathbb{R}\to S^1$, the derivative is \begin{align*} q'(t)=(-\sin t,\cos t). \end{align*} Thus \begin{align*} (q^*\alpha)_t(1) &=\alpha_{q(t)}(q'(t))\\ &=-(\sin t)(-\sin t)+(\cos t)(\cos t)\\ &=1. \end{align*} Therefore \begin{align*} q^*\alpha=dt. \end{align*} Since $\Omega^2(S^1)=0$, every $1$-form on $S^1$ is closed; in particular, $d\alpha=0$. [guided] We define the angular $1$-form by giving its value at each point of the circle. For $p=(p_1,p_2)\in S^1$ and $v=(v_1,v_2)\in T_pS^1\subset\mathbb{R}^2$, set \begin{align*} \alpha_p(v)=-p_2v_1+p_1v_2. \end{align*} This is smooth because it is the restriction to $S^1$ of the smooth ambient expression $-x_2\,dx_1+x_1\,dx_2$ on $\mathbb{R}^2$. Now compute its pullback along the parametrisation \begin{align*} q:\mathbb{R}&\to S^1\\ t&\mapsto(\cos t,\sin t). \end{align*} The derivative of $q$ at $t$ sends the tangent vector $1\in T_t\mathbb{R}\cong\mathbb{R}$ to \begin{align*} q'(t)=(-\sin t,\cos t)\in T_{q(t)}S^1. \end{align*} Therefore the pulled-back form satisfies \begin{align*} (q^*\alpha)_t(1) &=\alpha_{q(t)}(q'(t))\\ &=-(\sin t)(-\sin t)+(\cos t)(\cos t)\\ &=\sin^2 t+\cos^2 t\\ &=1. \end{align*} Since $dt$ is the standard $1$-form on $\mathbb{R}$ with $dt_t(1)=1$, this proves \begin{align*} q^*\alpha=dt. \end{align*} Finally, because $S^1$ is one-dimensional, $\Omega^2(S^1)=0$. The [exterior derivative](/theorems/1525) of a $1$-form is a $2$-form, so $d\alpha=0$. [/guided] [/step] [step:Show the angular form is not exact by its nonzero period] Suppose, for contradiction, that $\alpha=df$ for some smooth function \begin{align*} f:S^1\to\mathbb{R}. \end{align*} Then \begin{align*} dt=q^*\alpha=q^*(df)=d(f\circ q). \end{align*} Let $F:\mathbb{R}\to\mathbb{R}$ denote the smooth function $F=f\circ q$. The equality $dF=dt$ means $F'(t)=1$ for every $t\in\mathbb{R}$, hence \begin{align*} F(2\pi)-F(0)=\int_0^{2\pi}1\,d\mathcal{L}^1(t)=2\pi. \end{align*} But $q(2\pi)=q(0)$, so \begin{align*} F(2\pi)=f(q(2\pi))=f(q(0))=F(0), \end{align*} a contradiction. Therefore $\alpha$ is not exact, and $[\alpha]\neq 0$ in $H^1_{\mathrm{dR}}(S^1)$. [guided] To prove that $\alpha$ gives a nonzero cohomology class, we must rule out the possibility that $\alpha=df$ for a globally defined smooth function $f:S^1\to\mathbb{R}$. Assume such an $f$ exists. Pulling the equation $\alpha=df$ back along \begin{align*} q:\mathbb{R}\to S^1 \end{align*} gives \begin{align*} q^*\alpha=q^*(df). \end{align*} From the previous step, $q^*\alpha=dt$. Pullback commutes with exterior differentiation on functions, so $q^*(df)=d(f\circ q)$. Hence \begin{align*} d(f\circ q)=dt. \end{align*} Define \begin{align*} F:\mathbb{R}&\to\mathbb{R}\\ t&\mapsto f(q(t)). \end{align*} The equation $dF=dt$ says exactly that $F'(t)=1$ for all $t\in\mathbb{R}$. Integrating this derivative over the interval $[0,2\pi]$ gives \begin{align*} F(2\pi)-F(0)=\int_0^{2\pi}1\,d\mathcal{L}^1(t)=2\pi. \end{align*} However, the endpoints $0$ and $2\pi$ parametrize the same point of the circle: \begin{align*} q(0)=q(2\pi)=(1,0). \end{align*} Since $F=f\circ q$, this forces \begin{align*} F(2\pi)=f(q(2\pi))=f(q(0))=F(0), \end{align*} contradicting $F(2\pi)-F(0)=2\pi$. Therefore no global smooth primitive of $\alpha$ exists on $S^1$, so $[\alpha]\neq 0$. [/guided] [/step] [step:Subtract the average period from an arbitrary one-form] Let $\beta\in\Omega^1(S^1)$ be arbitrary. Since $\Omega^2(S^1)=0$, the form $\beta$ is closed. The pullback $q^*\beta\in\Omega^1(\mathbb{R})$ has a unique smooth coefficient function \begin{align*} b:\mathbb{R}\to\mathbb{R} \end{align*} such that \begin{align*} q^*\beta=b(t)\,dt. \end{align*} Because $q(t+2\pi)=q(t)$ and $q'(t+2\pi)=q'(t)$ for all $t\in\mathbb{R}$, the function $b$ is $2\pi$-periodic: \begin{align*} b(t+2\pi)=b(t). \end{align*} Define the real number \begin{align*} c:=\frac{1}{2\pi}\int_0^{2\pi} b(t)\,d\mathcal{L}^1(t). \end{align*} Define the smooth function \begin{align*} h:\mathbb{R}&\to\mathbb{R}\\ t&\mapsto \int_0^t \bigl(b(s)-c\bigr)\,d\mathcal{L}^1(s). \end{align*} Then $h'(t)=b(t)-c$. Moreover, \begin{align*} h(t+2\pi)-h(t) &=\int_t^{t+2\pi}\bigl(b(s)-c\bigr)\,d\mathcal{L}^1(s)\\ &=\int_0^{2\pi}\bigl(b(s)-c\bigr)\,d\mathcal{L}^1(s)\\ &=0, \end{align*} where the second equality uses the $2\pi$-periodicity of $b$. Hence $h$ is $2\pi$-periodic. [guided] Take an arbitrary smooth $1$-form \begin{align*} \beta\in\Omega^1(S^1). \end{align*} There are no nonzero $2$-forms on a one-dimensional manifold, so $d\beta=0$ automatically. We now show that the cohomology class of $\beta$ is a scalar multiple of $[\alpha]$. Pull $\beta$ back to the real line. Since every $1$-form on $\mathbb{R}$ is a smooth function times $dt$, there is a unique smooth function \begin{align*} b:\mathbb{R}\to\mathbb{R} \end{align*} such that \begin{align*} q^*\beta=b(t)\,dt. \end{align*} The function $b$ is $2\pi$-periodic because the parametrisation repeats after $2\pi$: both the base point and tangent vector agree, \begin{align*} q(t+2\pi)=q(t),\qquad q'(t+2\pi)=q'(t). \end{align*} Thus evaluating $\beta$ on these equal tangent vectors gives \begin{align*} b(t+2\pi)=b(t). \end{align*} The scalar multiple of $\alpha$ we must subtract is determined by the average period of $\beta$. Define \begin{align*} c:=\frac{1}{2\pi}\int_0^{2\pi} b(t)\,d\mathcal{L}^1(t). \end{align*} Then the function $b-c$ has zero integral over one full period. Define \begin{align*} h:\mathbb{R}&\to\mathbb{R}\\ t&\mapsto \int_0^t \bigl(b(s)-c\bigr)\,d\mathcal{L}^1(s). \end{align*} By the [fundamental theorem of calculus](/theorems/632) for smooth functions on $\mathbb{R}$, \begin{align*} h'(t)=b(t)-c. \end{align*} The zero-period condition makes $h$ periodic: \begin{align*} h(t+2\pi)-h(t) &=\int_t^{t+2\pi}\bigl(b(s)-c\bigr)\,d\mathcal{L}^1(s)\\ &=\int_0^{2\pi}\bigl(b(s)-c\bigr)\,d\mathcal{L}^1(s)\\ &=0. \end{align*} The second equality uses the $2\pi$-periodicity of $b-c$. This periodicity is the precise condition needed for $h$ to descend from $\mathbb{R}$ to a well-defined function on the quotient circle. [/guided] [/step] [step:Descend the zero-period primitive to the circle] Since $h$ is $2\pi$-periodic, there is a unique function \begin{align*} f:S^1&\to\mathbb{R} \end{align*} such that \begin{align*} f\circ q=h. \end{align*} The function $f$ is smooth because $q:\mathbb{R}\to S^1$ is a local diffeomorphism: on each open arc $V\subset S^1$ admitting a smooth inverse branch $\sigma:V\to\mathbb{R}$ of $q$, the restriction $f|_V$ is $h\circ\sigma$. We prove that $\beta-c\alpha=df$. Pulling back to $\mathbb{R}$ gives \begin{align*} q^*(\beta-c\alpha-df) &=q^*\beta-cq^*\alpha-d(f\circ q)\\ &=b(t)\,dt-c\,dt-h'(t)\,dt\\ &=\bigl(b(t)-c-(b(t)-c)\bigr)\,dt\\ &=0. \end{align*} The map $q$ is a local diffeomorphism and is surjective, so a $1$-form on $S^1$ whose pullback by $q$ is zero must itself be zero. Therefore \begin{align*} \beta-c\alpha=df. \end{align*} Thus every cohomology class in $H^1_{\mathrm{dR}}(S^1)$ is a scalar multiple of $[\alpha]$. [guided] Because $h(t+2\pi)=h(t)$ for every $t\in\mathbb{R}$, the value of $h(t)$ depends only on the point $q(t)\in S^1$, not on the chosen lift $t$. Hence there is a unique function \begin{align*} f:S^1\to\mathbb{R} \end{align*} satisfying \begin{align*} f\circ q=h. \end{align*} The function $f$ is smooth because $q:\mathbb{R}\to S^1$ is a local diffeomorphism: locally on $S^1$, one may choose a smooth inverse branch of $q$, and on such a branch $f$ is the composite of the smooth function $h$ with that inverse branch. We now check that $f$ is a primitive for $\beta-c\alpha$. Pull back the desired identity to $\mathbb{R}$: \begin{align*} q^*(\beta-c\alpha-df) &=q^*\beta-cq^*\alpha-q^*(df)\\ &=q^*\beta-cq^*\alpha-d(f\circ q). \end{align*} By definition of $b$, by the computation $q^*\alpha=dt$, and by $f\circ q=h$, this becomes \begin{align*} q^*(\beta-c\alpha-df) &=b(t)\,dt-c\,dt-dh\\ &=b(t)\,dt-c\,dt-h'(t)\,dt\\ &=\bigl(b(t)-c-(b(t)-c)\bigr)\,dt\\ &=0. \end{align*} Finally, since $q$ is a surjective local diffeomorphism, vanishing of the pullback detects vanishing of a $1$-form on $S^1$: for any $p\in S^1$ and $v\in T_pS^1$, choose $t\in\mathbb{R}$ with $q(t)=p$ and choose $a\in T_t\mathbb{R}$ with $dq_t(a)=v$; then the value of the form on $v$ is the value of its pullback on $a$. Therefore \begin{align*} \beta-c\alpha-df=0, \end{align*} so \begin{align*} \beta=c\alpha+df. \end{align*} This proves that the cohomology class $[\beta]$ equals $c[\alpha]$. [/guided] [/step] [step:Conclude that the angular class is a basis for first cohomology] The previous two steps show that $[\alpha]\neq 0$ and that every class in $H^1_{\mathrm{dR}}(S^1)$ is of the form $c[\alpha]$ for some $c\in\mathbb{R}$. Hence \begin{align*} H^1_{\mathrm{dR}}(S^1)=\operatorname{span}_{\mathbb{R}}\{[\alpha]\}\cong\mathbb{R}. \end{align*} Combining this with the computations in degrees $0$ and $k\ge 2$ gives \begin{align*} H^k_{\mathrm{dR}}(S^1)\cong \begin{cases} \mathbb{R}, & k=0,\\ \mathbb{R}, & k=1,\\ 0, & k\ge 2, \end{cases} \end{align*} and the generator in degree $1$ is the angular class $[\alpha]=[d\theta]$. [guided] We have proved two facts about degree $1$. First, the angular form $\alpha$ is closed and not exact, so its cohomology class satisfies \begin{align*} [\alpha]\neq 0 \end{align*} in $H^1_{\mathrm{dR}}(S^1)$. Second, for every smooth $1$-form $\beta\in\Omega^1(S^1)$, we constructed a real number \begin{align*} c=\frac{1}{2\pi}\int_0^{2\pi} b(t)\,d\mathcal{L}^1(t), \end{align*} where $q^*\beta=b(t)\,dt$, and a smooth function $f:S^1\to\mathbb{R}$ such that \begin{align*} \beta=c\alpha+df. \end{align*} Passing to cohomology classes kills the exact term $df$, so \begin{align*} [\beta]=c[\alpha]. \end{align*} Thus every class in $H^1_{\mathrm{dR}}(S^1)$ is a scalar multiple of $[\alpha]$, and $[\alpha]$ is nonzero. Therefore \begin{align*} H^1_{\mathrm{dR}}(S^1)=\operatorname{span}_{\mathbb{R}}\{[\alpha]\}\cong\mathbb{R}. \end{align*} Together with the earlier computation \begin{align*} H^0_{\mathrm{dR}}(S^1)\cong\mathbb{R} \end{align*} and the dimensional vanishing \begin{align*} H^k_{\mathrm{dR}}(S^1)=0\quad\text{for every }k\ge 2, \end{align*} we obtain \begin{align*} H^k_{\mathrm{dR}}(S^1)\cong \begin{cases} \mathbb{R}, & k=0,\\ \mathbb{R}, & k=1,\\ 0, & k\ge 2. \end{cases} \end{align*} The degree-$1$ generator is the angular class $[\alpha]$, which is the global meaning of the notation $[d\theta]$ on the circle. [/guided] [/step]