Theorem. Square tori have Betti numbers (1, 2, 1) [fdip-008Z]
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  • November 2, 2026

Let \(N_1\) and \(N_2\) be positive integers, and set \(V=N_1N_2\). The vertices are the \(V\) pairs \((x,y)\) of integers modulo \(N_1\) and \(N_2\), with addition of coordinates taken modulo those side lengths. For each vertex there is a horizontal edge \(e^h_{x,y}\) from \((x,y)\) to \((x+1,y)\), a vertical edge \(e^v_{x,y}\) from \((x,y)\) to \((x,y+1)\), and a face \(\sigma _{x,y}\). When a side length equals \(1\), each edge in that direction is a loop. The boundary of the face is \[\partial \sigma _{x,y} =e^h_{x,y}+e^v_{x+1,y}-e^h_{x,y+1}-e^v_{x,y}.\]

Real cochains are real functions on these cells. Write \(C^0\), \(C^1\) and \(C^2\) for the functions on vertices, edges and faces, of dimensions \(V\), \(2V\) and \(V\). A chain is a real linear combination of cells of one dimension, and the evaluation pairing \(\langle \beta ,c\rangle \) sums the pointwise products of a cochain with the coefficients of a chain. The coboundary \(d\) is the transpose of the boundary, in the sense that \(\langle d\beta ,c\rangle =\langle \beta ,\partial c\rangle \) whenever the degrees match. On vertex functions this is the incidence map of the ring-valued incidence: \((df)(e)=f(h(e))-f(s(e))\). In coordinates, \[\begin {aligned} (d_0f)^h(x,y)&=f(x+1,y)-f(x,y),\\ (d_0f)^v(x,y)&=f(x,y+1)-f(x,y),\\ (d_1\alpha )(\sigma _{x,y}) &=\alpha ^h(x,y)+\alpha ^v(x+1,y) -\alpha ^h(x,y+1)-\alpha ^v(x,y). \end {aligned}\]

The composite \(d_1\circ d_0\) is zero. Write \(b_0\) for the dimension of \(\ker d_0\), write \(b_1\) for the dimension of the quotient of \(\ker d_1\) by the image of \(d_0\), and write \(b_2\) for the dimension of the quotient of \(C^2\) by the image of \(d_1\). These Betti numbers are \(1\), \(2\) and \(1\).

Proof. 

  • November 2, 2026

Substitute the coordinate formulas. At the face \(\sigma _{x,y}\), \[\begin {aligned} (d_1d_0f)(\sigma _{x,y}) &=\bigl (f(x+1,y)-f(x,y)\bigr )\\ &\quad +\bigl (f(x+1,y+1)-f(x+1,y)\bigr )\\ &\quad -\bigl (f(x+1,y+1)-f(x,y+1)\bigr )\\ &\quad -\bigl (f(x,y+1)-f(x,y)\bigr ). \end {aligned}\] The 8 terms cancel in pairs, so \(d_1d_0f=0\). The pairing separates chains, and therefore the boundary of every face boundary is zero as well.

If \(d_0f=0\), then \(f(x+1,y)=f(x,y)\) and \(f(x,y+1)=f(x,y)\) for every vertex. Following the horizontal steps from \((0,0)\) and then the vertical steps gives \(f(x,y)=f(0,0)\). The kernel consists of the constant functions, so \(\dim \ker d_0=1\) and the rank of \(d_0\) is \(V-1\).

On a single cycle of length \(N\ge 1\), a real function \(g\) on the residues modulo \(N\) is the coboundary of a vertex function if and only if the sum of its values is zero. The forward difference of any vertex function sums to zero. Conversely, if that sum vanishes, set \(f(0)=0\) and let \(f(k)\) be the sum of \(g(0),\ldots ,g(k-1)\) for \(1\le k\le N-1\), with the empty sum equal to zero when \(N=1\). For \(0\le k\le N-2\) the forward difference equals \(g(k)\). The remaining difference is \(f(0)-f(N-1)\). It equals the negative of the sum of \(g(0),\ldots ,g(N-2)\), which the vanishing-sum hypothesis rewrites as \(g(N-1)\). When \(N=1\) both that difference and \(g(0)\) are zero. Thus \(g\) is a coboundary. The functions with vanishing difference are the constants, a one-dimensional space, so the coboundary on the cycle has rank \(N-1\).

Let \(\alpha \) satisfy \(d_1\alpha =0\). Write \(P^h(y)=\sum _x\alpha ^h(x,y)\) and \(P^v(x)=\sum _y\alpha ^v(x,y)\). Sum the face equation over \(x\). The vertical terms cancel because \(x+1\) runs through every residue, and the result is \(P^h(y)=P^h(y+1)\). Sum the face equation over \(y\). The horizontal terms cancel, and the result is \(P^v(x)=P^v(x+1)\). Thus both quantities are constants; call them \(P^h\) and \(P^v\). If \(\alpha =d_0f\), each is a sum of a telescoping difference, so \((P^h,P^v)=(0,0)\).

Conversely, suppose \(d_1\alpha =0\) and \((P^h,P^v)=(0,0)\). For each fixed \(y\), the horizontal row sums to zero, so the cycle criterion supplies a function \(f(\cdot ,y)\) with \(f(0,y)=0\) and horizontal difference \(\alpha ^h\). Define \(\delta (x,y)=\alpha ^v(x,y)-\bigl (f(x,y+1)-f(x,y)\bigr )\). Its horizontal difference is \[\alpha ^v(x+1,y)-\alpha ^v(x,y) -\bigl (\alpha ^h(x,y+1)-\alpha ^h(x,y)\bigr ).\] The face equation rearranges to the vanishing of that expression. Hence \(\delta (x,y)\) depends only on \(y\). The normalization \(f(0,y)=0\) gives the common value \(k(y)=\alpha ^v(0,y)\). The sum of \(k\) is \(P^v=0\), so the cycle criterion supplies \(c\) with \(c(y+1)-c(y)=k(y)\). The function \(f(x,y)+c(y)\) then has horizontal difference \(\alpha ^h\) and vertical difference \(\alpha ^v\). Every cocycle with vanishing periods is therefore a coboundary.

The identity \(d_1d_0=0\) puts the image of \(d_0\) inside \(\ker d_1\). Coboundaries have vanishing periods, and every cocycle with vanishing periods is a coboundary, so that image is exactly the set of cocycles whose periods vanish. The edge field with every horizontal value \(1\) and every vertical value \(0\) has coboundary zero on each face, because the face sees \(+1\) on its lower edge and \(-1\) on its upper edge. Its periods are \((N_1,0)\). The field with the two roles exchanged is likewise a cocycle, with periods \((0,N_2)\). For real \(a\) and \(b\), the combination with horizontal value \(a/N_1\) and vertical value \(b/N_2\) is a cocycle of periods \((a,b)\). The period map on \(\ker d_1\) is therefore a surjective linear map to \(\mathbb R^2\), with kernel equal to the image of \(d_0\). Rank nullity gives \[\dim \ker d_1=(V-1)+2=V+1,\] and the quotient dimension is \[b_1=(V+1)-(V-1)=2.\] The rank of \(d_1\) is \(2V-(V+1)=V-1\).

Summing a coboundary over every face cancels: each horizontal value appears once with each sign as \(y\) runs through a full set of residues, and each vertical value does the same as \(x\) does. The sum of the face values is therefore a linear functional vanishing on the image of \(d_1\). The constant function \(1\) has sum \(V\ne 0\), so the functional is nonzero, its kernel has dimension \(V-1\), and that constant is not a coboundary. The image already has dimension \(V-1\), so the image equals the kernel of the sum. Therefore \(b_2=1\).

The constant vertex functions span the degree-zero cohomology. The two edge fields just named have linearly independent periods, and the degree-one quotient is two-dimensional, so their classes form a basis: one winds around the first coordinate and the other winds around the second. The constant face functions span the degree-two cohomology.