Theorem. The cup product obeys graded Leibniz [fdip-0093]
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  • November 2, 2026

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. 

  • 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.