answer:To compute the dimension of the space of harmonic 1-forms on a connected and smooth closed surface with genus g, we will use De Rham cohomology and the Hodge decomposition theorem. The Hodge decomposition theorem states that for a compact Riemannian manifold M, the space of differential forms Ω^k(M) can be decomposed into the direct sum of three subspaces: the space of exact forms, the space of coexact forms, and the space of harmonic forms. In other words, Ω^k(M) = dΩ^(k-1)(M) ⊕ δΩ^(k+1)(M) ⊕ H^k(M), where H^k(M) is the space of harmonic k-forms. For a connected and smooth closed surface with genus g, we are interested in the space of harmonic 1-forms, H^1(M). By the Hodge decomposition theorem, we know that H^1(M) is isomorphic to the first De Rham cohomology group, H^1(M; R). The dimension of H^1(M; R) is given by the first Betti number, b_1(M). Now, let's compute b_1(M) for a surface of genus g. Recall that the Betti numbers are topological invariants, and they can be computed using the Euler characteristic. For a connected and smooth closed surface with genus g, the Euler characteristic is given by χ(M) = 2 - 2g. The Betti numbers satisfy the relation χ(M) = b_0(M) - b_1(M) + b_2(M). Since M is connected, b_0(M) = 1, and since M is a closed surface, b_2(M) = 1. Therefore, we have: 2 - 2g = 1 - b_1(M) + 1 b_1(M) = 2g Thus, the dimension of the space of harmonic 1-forms on a connected and smooth closed surface with genus g is 2g.

answer:We will use the Morse theory to prove the statement. Let f: X to mathbb{R} be a Morse function on the compact and connected manifold X. Recall that a Morse function is a smooth function with non-degenerate critical points, i.e., the Hessian matrix at each critical point is non-singular. By Sard's theorem, we know that almost every smooth function is a Morse function, so we can assume that f is Morse without loss of generality. Let m_k denote the number of critical points of f with index k, where the index of a critical point is the number of negative eigenvalues of the Hessian matrix at that point. The Morse inequalities state that for each k = 0, 1, dots, n (where n is the dimension of X), we have: m_k geq dim H^k(X, mathbb{R}). Now, we know that H^1(X, mathbb{R}) is one-dimensional, so dim H^1(X, mathbb{R}) = 1. Thus, we have: m_1 geq 1. Next, we consider the Euler characteristic chi(X). Recall that the Euler characteristic is defined as the alternating sum of the Betti numbers, which are the dimensions of the homology groups: chi(X) = sum_{k=0}^n (-1)^k dim H_k(X, mathbb{R}). By Poincaré duality, we have dim H_k(X, mathbb{R}) = dim H^{n-k}(X, mathbb{R}) for all k. Therefore, we can rewrite the Euler characteristic as: chi(X) = sum_{k=0}^n (-1)^k dim H^k(X, mathbb{R}). Now, let c = min_{0 leq k leq n} frac{m_k}{dim H^k(X, mathbb{R})}, where we take the minimum over all non-zero dimensions. Note that c > 0 since all dimensions are non-zero. Then, we have: c cdot chi(X) = c cdot sum_{k=0}^n (-1)^k dim H^k(X, mathbb{R}) leq sum_{k=0}^n (-1)^k m_k. Since m_1 geq 1, we have: sum_{k=0}^n (-1)^k m_k geq m_0 - m_1 + sum_{k=2}^n (-1)^k m_k geq m_0 - 1 + sum_{k=2}^n (-1)^k m_k. Thus, we have: c cdot chi(X) leq sum_{k=0}^n (-1)^k m_k leq m_0 - 1 + sum_{k=2}^n (-1)^k m_k. Since the number of critical points of f is given by sum_{k=0}^n m_k, we have: c cdot chi(X) leq sum_{k=0}^n m_k - 1. Adding 1 to both sides, we obtain: c cdot chi(X) + 1 leq sum_{k=0}^n m_k. Hence, there exists a constant c in mathbb{R} such that the number of critical points of f is greater than or equal to c cdot chi(X).

answer:To compute the De Rham cohomology groups H^k_{DR}(M) for k=0,1,dots, n, we first need to understand the De Rham complex of differential forms on M. The De Rham complex is given by the sequence of spaces of differential forms and their exterior derivatives: 0 rightarrow Omega^0(M) xrightarrow{d} Omega^1(M) xrightarrow{d} cdots xrightarrow{d} Omega^n(M) rightarrow 0 Here, Omega^k(M) denotes the space of smooth k-forms on M, and d is the exterior derivative, which takes a k-form to a (k+1)-form. The De Rham cohomology groups H^k_{DR}(M) are defined as the quotient of the space of closed k-forms (i.e., those whose exterior derivative is zero) by the space of exact k-forms (i.e., those that are the exterior derivative of a (k-1)-form). In other words: H^k_{DR}(M) = frac{ker(d: Omega^k(M) rightarrow Omega^{k+1}(M))}{operatorname{im}(d: Omega^{k-1}(M) rightarrow Omega^k(M))} Now, let's compute the De Rham cohomology groups for each k: 1. For k=0, we have: H^0_{DR}(M) = frac{ker(d: Omega^0(M) rightarrow Omega^1(M))}{operatorname{im}(d: Omega^{-1}(M) rightarrow Omega^0(M))} Since there are no (-1)-forms, the denominator is trivial. The kernel of d: Omega^0(M) rightarrow Omega^1(M) consists of constant functions, as their exterior derivative is zero. Therefore, H^0_{DR}(M) is isomorphic to the space of constant functions on M. Since M is connected, this space is isomorphic to mathbb{R}. 2. For k=n, we have: H^n_{DR}(M) = frac{ker(d: Omega^n(M) rightarrow Omega^{n+1}(M))}{operatorname{im}(d: Omega^{n-1}(M) rightarrow Omega^n(M))} Since there are no (n+1)-forms, the kernel is trivial. Therefore, H^n_{DR}(M) is isomorphic to the space of closed n-forms modulo exact n-forms. By the Poincaré lemma, every closed n-form is locally exact. However, since M is compact, a global argument is needed. By the Stokes' theorem, the integral of an exact n-form over M is zero. If the integral of a closed n-form is nonzero, it cannot be exact. Therefore, H^n_{DR}(M) is isomorphic to mathbb{R} if M has a nonzero volume form, and is trivial otherwise. 3. For 1 leq k leq n-1, the De Rham cohomology groups H^k_{DR}(M) depend on the topology of M. In general, they can be computed using the Mayer-Vietoris sequence, the Künneth formula, or other algebraic topology tools.

answer:The statement is False. Proof by contradiction: Suppose there exists a closed 1-form beta on M such that eta = dbeta and alpha = [beta] in H^2_{dR}(M). Since eta is not exact, there is no 0-form gamma on M such that eta = dgamma. However, by our assumption, eta = dbeta, which contradicts the fact that eta is not exact. Furthermore, since alpha wedge eta = 0, we have that alpha wedge dbeta = 0. Since alpha is a closed 2-form, dalpha = 0. Now, if alpha = [beta] in H^2_{dR}(M), then alpha and beta are cohomologous, which means there exists a 1-form delta such that alpha - beta = ddelta. But this would imply that alpha wedge dbeta = ddelta wedge dbeta = 0, which contradicts the fact that alpha wedge eta = 0. Therefore, the statement is false, and there does not exist a closed 1-form beta on M such that eta = dbeta and alpha = [beta] in H^2_{dR}(M).