M2 Wess Zumino Family Curvature And Flat Ambiguity

The forms $dx,dy$ and $\sigma$ are closed, so $d\omega =0$. The product $S=T^2\times S^2$ is a nonempty compact connected oriented smooth four-manifold without boundary; [F3] gives $H_4(S;\mathbb{Z})=\mathbb{Z}[S]$. Triangulate $S$ by [F5], order its finitely many vertices, and sum its oriented smooth top simplices to obtain a smooth integral cycle $t$. Identical parametrized interior faces cancel, and its local coefficient at an interior point of any top simplex is $+1$, so $[t]=[S]$. By [F4], every smooth integral four-cycle $z$ satisfies $z-nt=\mathrm{\partial }b$ for some integer $n$ and smooth integral five-chain $b$. By [F6], ${\int }_z\omega =n{\int }_t\omega +{\int }_{b}d\omega =n$, because ${\int }_t\omega ={\int }_S\omega =({\int }_{C}dx\wedge dy)({\int }_{S^2}\sigma )=1$. Thus the required periods on all smooth integral cycles are integers, without any cellular-to-singular inference.

For $N\ge 0$, put $K_N(t)=(N+1{)}^{-1}\mathrm{\mid }{\sum }_{j=0}^N{e}^{2\pi ijt}{\mathrm{\mid }}^2$. Expanding the square gives $K_N={\sum }_{\mathrm{\mid }k\mathrm{\mid }\le N}(1-\mathrm{\mid }k\mathrm{\mid }\mathrm{/}(N+1))e^{2\pi ikt}$, hence $K_N\ge 0$ and ${\int }_0^1{K}_N=1$ by direct integration of exponentials. For $\delta \le \mathrm{dist}(t,\mathbb{Z})\le 1\mathrm{/}2$, the geometric-sum formula gives $K_N(t)\le ((N+1){\sin }^2(\pi \delta ){)}^{-1}$. Thus the product kernel $K_N(s)K_N(t)$ has mass at most $2\mathrm{/}((N+1){\sin }^2(\pi \delta ))$ outside the square where both circle distances are less than $\delta$. For a continuous periodic $u$, convolution is a trigonometric polynomial and its error is at most the modulus of continuity of $u$ on that square plus $2\mathrm{\parallel }u{\mathrm{\parallel }}_{\mathrm{\infty }}$ times this tail mass. First shrink $\delta$ and then increase $N$ to obtain uniform approximation.

By [F2] there is an ordinary degree-four differential character on $S$ with curvature $\omega$. Also $q_{!}\omega =dx\wedge dy$, since the integral of $\sigma$ over each fibre is $1$. On a chart with lift $x_i$, $d{c}_i=2\pi dx\wedge dy\wedge \sigma =2\pi \omega$ and $q_{!}{c}_i=2\pi x_{i}dy$. Therefore $d{A}_i=2\pi dx\wedge dy$, in agreement with [F1].

Finite unions of half-open rectangles approximate every Lebesgue measurable subset of $[0,1{)}^2$ in measure: the sets approximable in symmetric-difference measure form a sigma algebra containing the rectangles. For countable unions first truncate to a finite union, using continuity from below of the finite measure, and then approximate its finitely many terms; complements preserve the approximation. Hence this class contains the Borel sigma algebra, and completion adds only sets equal modulo null sets to Borel sets. Rectangle indicators are approximated in $L^2$ by continuous periodic ramps, bounded between zero and one and differing only in strips of area tending to zero along their boundaries. Truncating an $L^2$ function and quantizing its bounded real and imaginary parts approximates it by simple measurable functions, since the discarded $L^2$ tail tends to zero and the quantization error is uniformly small. Approximate their finitely many indicator sets by rectangles and ramps to obtain density of continuous periodic functions in $L^2$. Step 1.2 therefore proves that the orthonormal exponentials $e_{mn}=e^{2\pi i(mx+ny)}$ have dense finite span. Orthogonal finite projections then converge in $L^2$ and give Parseval's identity.

For the $x$ identification from the fibre at $(x+1,y)$ to that at $(x,y)$, put $g(y)=e^{-2\pi iy}$ and $B(x)=-2\pi ixdy$. Then $g^{-1}dg=-2\pi idy$ and $B(x+1)=B(x)+g^{-1}dg$, so $(d+B(x))(gu)=g(d+B(x+1))u$. Thus the identification preserves the connection and norm. The $y$ identification preserves $B$; it commutes with the $x$ identification because $g(y+1)=g(y)$. The free properly discontinuous base lattice action and these smooth linear fibre maps give local quotient charts on small rectangles, hence a smooth Hermitian line with connection. Its curvature is $F_{\mathrm{\nabla }}=dB=-2\pi idx\wedge dy$, and the convention $c_1=(i\mathrm{/}2\pi )F_{\mathrm{\nabla }}$ gives degree $1$.

Write $u=\sum a_{mn}{e}_{mn}$ in this complete basis. Testing the weak derivative definition on exponentials shows that the derivative coefficients are $2\pi im{a}_{mn}$ and $2\pi in{a}_{mn}$. Parseval gives $u\in H^1$ only if $\sum (1+4{\pi }^2(m^2+n^2))\mathrm{\mid }{a}_{mn}{\mathrm{\mid }}^2<\mathrm{\infty }$. Conversely, that weighted summability makes finite rectangular partial sums and both derivatives converge in $L^2$. Passing to the limit in their integration-by-parts identities against each smooth periodic test function proves the weak derivative identities, hence membership in $H^1$ and the asserted norm formula. These weighted sequence spaces are complete, so this also proves that the specified $H^1$ is a Hilbert space. Direct differentiation on finite sums and boundedness $H^1\to L^2$ show that ${\bar{\mathrm{\partial }}}_{\theta }$ has multiplier ${\lambda }_{mn}(\theta )=\pi (i(m-\theta )-n)$ on the whole domain.

Since $d(dx)=0$, $F_{{\mathrm{\nabla }}_{\theta }}=0$ for each $\theta$. Parallel transport along the x-cycle solves $s^{\mathrm{\prime }}=2\pi i\theta s$, so its holonomy is $e^{2\pi i\theta }$. Thus $F_0$ and $F_{1\mathrm{/}2}$ have equal zero curvature and degree zero but distinct holonomy; tensoring either with the degree-one line leaves its curvature and degree unchanged.

For $\theta =0$, ${\lambda }_{mn}$ vanishes exactly at $(m,n)=(0,0)$. On all other modes, $\mathrm{\mid }{\lambda }_{mn}{\mathrm{\mid }}^2\ge ({\pi }^2\mathrm{/}16)(1+m^2+n^2)$; the same bound holds on every mode for $\theta =1\mathrm{/}2$, since $\mathrm{\mid }m-1\mathrm{/}2\mathrm{\mid }\ge 1\mathrm{/}2$ and $m^2\le 2(m-1\mathrm{/}2{)}^2+1\mathrm{/}2$. Division by the nonzero multipliers therefore sends any square-summable target sequence to the weighted domain of step 3.2. For $\theta =0$ the range is exactly the closed codimension-one space of targets with zero constant coefficient; its kernel consists exactly of constants. For $\theta =1\mathrm{/}2$ the map is bijective, so has closed full range and zero kernel and cokernel. Both operators are Fredholm, with indices $1-1=0$ and $0-0=0$, but their kernel dimensions are one and zero. These kernel representatives are smooth by their explicit Fourier coefficients. They are mathematical Dolbeault zero modes, not physical vacuum states.

The local transgression computes curvature and leaves the two flat holonomies free. The explicit equal-degree pair has equal net index and different zero-mode spaces, so curvature or degree alone does not determine the individual modes.