Dirac–Kähler operators [fdip-0091]
✍️sourceAGENTDRAFTED
- November 2, 2026
Dirac–Kähler operators [fdip-0091]
✍️sourceAGENTDRAFTED
- November 2, 2026
The centered Dirac operator on a periodic grid has kernel dimension \(2^{d+e}\), where \(e\) counts the even periods. On a square torus that dimension changes when a side length changes parity. The operator built from the coboundary does not. On every square torus the kernel of \(d+d^\dagger \) has dimension \(4\).
The exterior algebra of a free module carries a multiplication and a contraction for each basis vector. No quadratic form enters those two operations. Their ordered products are every linear endomorphism of the algebra, and the sums of matching pairs obey the Clifford relation of the standard dot product. The following section proves both statements. The section after it takes up the product of cochains.
Theorem 1. Ordered exterior operators span the endomorphisms [fdip-0092]AGENTDRAFTED
- November 2, 2026
Theorem 1. Ordered exterior operators span the endomorphisms [fdip-0092]AGENTDRAFTED
- November 2, 2026
On every square torus the kernel of \(d+d^\dagger \) has dimension \(4\), by the Hodge theorem. For periods at least \(3\), the centered Dirac operator has kernel dimension \(2^{2+e}\). That dimension equals \(4\) when both periods are odd, and it is strictly larger when either period is even.
Let \(k\) be a commutative ring and let \(n\) be a positive integer. Write \(\bigwedge \) for the exterior algebra of the free module on basis vectors \(e_1,\ldots ,e_n\), with basis the monomials \(e_I\) indexed by subsets \(I\), written in increasing order, and with \(e_\varnothing =1\). Exterior multiplication and contraction along the basis are the linear operators \[\epsilon _i(e_I)=e_i\wedge e_I,\] \[\iota _i(e_I)=0\] when \(i\notin I\), and, when \(i\) occupies position \(p\) in \(I\), \[\iota _i(e_I)=(-1)^p e_{I\setminus \{i\}}.\] For subsets \(S\) and \(T\), write \(\epsilon _S\) and \(\iota _T\) for the compositions of these operators in increasing index order, the largest index acting first, and write \(M_{S,T}=\epsilon _S\circ \iota _T\). There are \(2^n\) choices for each subset, hence \(4^n\) such operators, and they are a basis of the endomorphism ring of \(\bigwedge \). In particular \(\gamma _i=\epsilon _i+\iota _i\) satisfies \[\gamma _i\gamma _j+\gamma _j\gamma _i=2\delta _{ij}I.\] The configuration Clifford algebra is the square quotient of a supplied quadratic form. The operators here are defined before any such form, and the displayed relation is the square relation of the standard dot product. The note does not claim that the resulting Clifford representation is injective.
Proof.
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Proof.
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The dimension \(4\) is the kernel dimension in the Hodge theorem. The centered count is the cited formula at \(d=2\). It equals \(4\) when \(e=0\). When \(e\ge 1\) the factor \(2^e\) makes it strictly larger. The two statements have the scopes already named.
The monomials are a basis, so operator identities may be checked on \(e_I\). If \(i\in I\), then \(e_i\wedge e_I=0\). If \(i\notin I\) and \(q\) indices of \(I\) are smaller than \(i\), then \(e_i\wedge e_I=(-1)^q e_{I\cup \{i\}}\). Thus \(\epsilon _i^2=0\). For \(i\ne j\) the wedge of the two basis vectors changes sign when their order is exchanged, so \(\epsilon _i\epsilon _j+\epsilon _j\epsilon _i=0\). A monomial contains \(i\) at most once, so \(\iota _i^2=0\).
For the contractions, take \(i\ne j\) and suppose both lie in \(I\), with positions \(p\) and \(r\) and with \(i<j\), so \(p<r\). Removing \(j\) first contributes \((-1)^r\), and \(i\) is still at position \(p\), so the second removal contributes \((-1)^p\). Removing \(i\) first contributes \((-1)^p\), after which \(j\) stands one place earlier, and the second removal contributes \((-1)^{r-1}\). The two signs are opposite, and the remaining monomial is the same. If either index is missing, both double contractions vanish. Hence \(\iota _i\iota _j+\iota _j\iota _i=0\).
Now compare \(\iota _j\epsilon _i\) with \(\epsilon _i\iota _j\). If \(i\ne j\) and \(j\notin I\), the contraction of the input is zero. Exterior multiplication either vanishes or produces a monomial that still omits \(j\), so the later contraction vanishes as well. If \(j\in I\) and \(i\in I\), exterior multiplication of the input vanishes, and the contraction still leaves \(i\) in place, so the later exterior multiplication vanishes. The remaining pattern is \(j\in I\) and \(i\notin I\). Let \(p\) be the position of \(j\) and let \(q\) be the number of indices of \(I\) smaller than \(i\). Removing \(j\) contributes \((-1)^p\). If \(i<j\), wedging \(e_i\) back contributes the same sign \((-1)^q\) that \(e_i\wedge e_I\) carries, while the later removal of \(j\) from \(I\cup \{i\}\) contributes \((-1)^{p+1}\), because \(i\) stands before \(j\). If \(i>j\), wedging \(e_i\) after the removal contributes \((-1)^{q-1}\), because \(j\) was one of the indices counted by \(q\), while the removal of \(j\) from \(I\cup \{i\}\) still contributes \((-1)^p\). In both subcases the two compositions land on the same monomial with opposite signs. If \(i=j\) and \(i\notin I\), the contraction of the input vanishes and \(\iota _i\epsilon _i\) returns \(e_I\), and the product of the two signs \((-1)^q\) is \(1\). If \(i\in I\), exterior multiplication of the input vanishes and \(\epsilon _i\iota _i\) returns \(e_I\), and the product of the two signs \((-1)^p\) is \(1\). Therefore \(\epsilon _i\iota _j+\iota _j\epsilon _i=\delta _{ij}I\).
The square of \(\gamma _i\) is \(\epsilon _i\iota _i+\iota _i\epsilon _i=I\). For \(i\ne j\) each of the 4 pairs in the expansion of \(\gamma _i\gamma _j+\gamma _j\gamma _i\) anticommutes, so the sum is zero. This is the displayed Clifford relation.
The monomials are a free basis, so the endomorphism ring is free on the operators that send one monomial to one monomial and kill the rest. Order the pairs of subsets so that the cardinality of \(T\) increases, and index both the operators \(M_{S,T}\) and these monomial matrix units by that order. The coefficient of \(e_S\) in \(M_{S',T'}e_T\) can be nonzero only when \(T'\subseteq T\). A column with a strictly larger second set therefore does not meet this row. When the cardinalities are equal, necessarily \(T'=T\). The contraction \(\iota _T\) removes every factor of \(e_T\). Each removal meets a present factor, so the image of the contraction is \((\pm 1)\,1\). The following exterior product inserts the factors in \(S'\), each into a monomial that does not yet contain it, and the image is \((\pm 1)\,e_{S'}\). The coefficient of \(e_S\) is therefore zero unless \(S'=S\), in which case it is \(\pm 1\). Within one cardinality the block is diagonal. A strictly smaller \(T'\) meets the row in an earlier column. The matrix is consequently lower triangular, with units \(\pm 1\) on the diagonal. Given a target endomorphism, its coefficients against the \(M_{S,T}\) are the unique solution of that system: start from the empty second set, invert the diagonal unit, and at each later row use the coefficients already found. The operators are therefore a basis.
For a single basis vector the 4 operators are the identity, exterior multiplication, contraction, and their composite. On a space of dimension \(2\) any 4 independent endomorphisms are a basis, and the triangular count supplies the independence.
Theorem 2. The cup product obeys graded Leibniz [fdip-0093]AGENTDRAFTED
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Theorem 2. The cup product obeys graded Leibniz [fdip-0093]AGENTDRAFTED
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Let the vertices of a finite simplicial complex be ordered, and let a cochain take values in a commutative ring. On an increasing simplex the cup product of a cochain \(\alpha \) of degree \(k\) and a cochain \(\beta \) of degree \(l\) reads the front face of dimension \(k\) and the back face of dimension \(l\): \[(\alpha \cup \beta )(v_0,\ldots ,v_{k+l}) =\alpha (v_0,\ldots ,v_k)\,\beta (v_k,\ldots ,v_{k+l}).\] The coboundary is the alternating sum of the faces with one vertex removed. Then \[d(\alpha \cup \beta )=(d\alpha )\cup \beta +(-1)^k\alpha \cup (d\beta ).\] The same formula does not make the product graded-commutative on cochains.
Let \(G\) be a group and let the transport of a path be the one in the plaquette convention: the first edge stands on the right, and a reversed edge contributes the inverse. On the rectangle with \(M\) squares horizontally and \(N\) squares vertically, label the square whose lower-left corner is \((i,j)\) by \(p_{i,j}\). Let \(\gamma _{i,j}\) run from the lower-left corner of the rectangle along the bottom to column \(i\) and then up to row \(j\). The boundary holonomy, traversed counterclockwise from that corner, equals \[\prod _{i=0}^{M-1}\prod _{j=N-1}^{0} U_{\gamma _{i,j}}^{-1}U_{p_{i,j}}U_{\gamma _{i,j}},\] with the left column on the left of the product and, inside a column, the top square on the left.
Proof.
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Proof.
- November 2, 2026
Evaluate both sides of the Leibniz identity on an ordered simplex of dimension \(k+l+1\), with vertices \(v_0\) through \(v_{k+l+1}\). The coboundary of the product is the alternating sum of the products on the faces. On the face that omits \(v_i\) with \(i\le k\), the front face of dimension \(k\) is \((v_0,\ldots ,\widehat {v_i},\ldots ,v_{k+1})\) and the back begins at \(v_{k+1}\). Those are the terms \(i\le k\) of \((d\alpha )\cup \beta \). On the face that omits \(v_i\) with \(i\ge k+1\), write \(j=i-k\). The front is \((v_0,\ldots ,v_k)\) and the back omits the vertex of index \(j\) in \((v_k,\ldots ,v_{k+l+1})\). The sign \((-1)^{k+j}\) is the sign in \((-1)^k\alpha \cup (d\beta )\). The two sums therefore share every face except the face \(i=k+1\) of \((d\alpha )\cup \beta \) and the face \(j=0\) of \((-1)^k\alpha \cup (d\beta )\). Those two faces are the same product \(\alpha (v_0,\ldots ,v_k)\beta (v_{k+1},\ldots ,v_{k+l+1})\), with signs \((-1)^{k+1}\) and \((-1)^k\), which add to zero.
On the simplex \((0,1,2)\), let \(\alpha \) and \(\beta \) be \(1\)-cochains with \(\alpha (0,1)=1\), \(\alpha (1,2)=0\), \(\beta (0,1)=0\) and \(\beta (1,2)=1\). Then \((\alpha \cup \beta )(0,1,2)=1\) and \((\beta \cup \alpha )(0,1,2)=0\). The two products differ as cochains. This note does not prove that their difference becomes a coboundary after both factors are closed.
Write \(U_{p}\) for the plaquette holonomy. For the square with lower-left corner \((i,j)\), if \(b\) is its bottom edge, \(r\) its right edge, \(t\) its top edge and \(s\) its left edge, each oriented rightward or upward, then \(U_p=s^{-1}t^{-1}rb\). The displayed product is proved by adding columns from the left. A single column has bottom edge \(b_0\), right edges \(r_0,\ldots ,r_{N-1}\) from bottom to top, top edge \(t\), and left edges \(s_0,\ldots ,s_{N-1}\). Its boundary transport is \(s_0^{-1}\cdots s_{N-1}^{-1}t^{-1}r_{N-1}\cdots r_0b_0\). The path \(\gamma _{0,j}\) is the left side up to height \(j\), with transport \(\sigma _j=s_{j-1}\cdots s_0\) and \(\sigma _0=1\). One square is already the boundary word. For several squares the conjugate by \(\sigma _j^{-1}\) is \(\sigma _j^{-1}s_j^{-1}t_j^{-1}r_jb_j\sigma _j\). In the product from the top of the column down to the bottom, the junction of the square at height \(j\) with the square below it is \(\sigma _j\sigma _{j-1}^{-1}s_{j-1}^{-1}=1\), because \(\sigma _j=s_{j-1}\sigma _{j-1}\), and \(b_jt_{j-1}^{-1}=1\), because the bottom of the upper square is the top of the square below it. The right edges remain in order from the top down, and the surviving left edge is the full inverse \(s_0^{-1}\cdots s_{N-1}^{-1}\). The product is the boundary word.
Suppose the identity holds for the rectangle with \(M-1\) columns, and attach one column on the right. The conjugating paths of the columns already present do not use the new column, so the inductive product is the old boundary transport times the new column product, the new factor standing on the right. Let \(\beta \) be the transport along the old bottom, \(S\) the transport up the shared vertical side, \(T\) the reversed old top and \(L\) the reversed left side. The old boundary transport is \(LTS\beta \). The new boundary inserts the new bottom edge \(b_0\) after \(\beta \), replaces the shared side by the new right side \(R\), and inserts the inverse of the new top edge \(t\) before \(T\). Its transport is \(LTt^{-1}Rb_0\beta \). Each new path is the old bottom followed by the upward path of the previous paragraph, so each new conjugate is \(\beta ^{-1}\) times the corresponding column factor times \(\beta \). Between successive squares \(\beta \) meets \(\beta ^{-1}\), and the product of those conjugates is \(\beta ^{-1}S^{-1}t^{-1}Rb_0\beta \). Here \(S\) is the upward transport of the shared side, lowest edge on the right, so \(\beta ^{-1}S^{-1}\) is the inverse of the path along the old bottom and then up that side. Multiplying on the left by \(LTS\beta \) cancels \(\beta \) against \(\beta ^{-1}\) and \(S\) against \(S^{-1}\), and leaves the new boundary transport. Finiteness of the group is not used.