Modular Categories, Verlinde Formula, and TQFT

On homogeneous simple grades, the associator and braiding are multiplication by the scalars $\omega$ and $r$. Therefore the two paths around the monoidal pentagon agree exactly by the displayed pentagon equation, and each braided hexagon diagram agrees exactly by its corresponding displayed equation. These equalities extend to finite direct sums by linearity and to all homogeneous morphisms by naturality of scalar maps. The grade $0$ acts as the tensor unit because normalization makes every unit associator and braiding scalar equal to $1$.

The chosen evaluation and coevaluation have the stated sources and targets. The first snake has scalar $\omega (g,-g,g{)}^{-1}\omega (g,-g,g)=1$. The second has scalar $\omega (g,-g,g{)}^{-1}\omega (-g,g,-g{)}^{-1}=1$, because the pentagon at $(g,-g,g,-g)$ gives $\omega (g,-g,g)\omega (-g,g,-g)=1$. Hence each grade has the specified left dual.

The scalar $b(g,h)=r(g,h)r(h,g)$ is the double-braiding value. The displayed formula for $q(g+h)$ is exactly the balancing identity on two homogeneous simples, while $q(-g)=q(g)$ is the dual-compatibility condition. Thus the stated scalar twist is ribbon-compatible; the pivotal dimension-one convention is an explicit part of the input data.

For an integer $r$ with $0<\mathrm{\mid }r\mathrm{\mid }<2\mathrm{\ell }$, put $C_r={\sum }_{m=1}^{\mathrm{\ell }-1}\cos (rm\pi \mathrm{/}\mathrm{\ell })$. If $r$ is even, then $z=e^{2\pi i(r\mathrm{/}2)\mathrm{/}\mathrm{\ell }}$ is a nontrivial $\mathrm{\ell }$-th root of unity; the geometric sum ${\sum }_{m=0}^{\mathrm{\ell }-1}{z}^m=0$ has real part $1+C_r$, so $C_r=-1$. If $r$ is odd, the geometric sum of $e^{i\pi rm\mathrm{/}\mathrm{\ell }}$ for $0\le m<2\mathrm{\ell }$ is zero. Pairing terms at $m$ and $2\mathrm{\ell }-m$ for $1\le m\le \mathrm{\ell }-1$ leaves twice $C_r$; the two unpaired real terms are $1+\cos (r\pi )=0$, so $C_r=0$. Also $C_0=\mathrm{\ell }-1$.

The product-to-sum identity gives the left side as $\frac{1}{2}(C_{a-b}-C_{a+b})$, with $C_0$ used when $a=b$. If $a=b$, then $a+b$ is nonzero even and less than $2\mathrm{\ell }$, so the value is $\frac{1}{2}((\mathrm{\ell }-1)-(-1))=\mathrm{\ell }\mathrm{/}2$. If $a\mathrm{\ne }b$ and $a,b$ have the same parity, both nonzero indices $a-b$ and $a+b$ are even and their $C$ values are $-1$; if they have opposite parity, both indices are odd and their values are $0$. In both off-diagonal cases the difference is zero, proving the formula.

For a basis element $y$, form $P_{\chi }={\sum }_{x\in I}\chi (x)x^{\ast }$ in $A{\otimes }_{\mathbb{Z}}\mathbb{C}$. The coefficient of $z$ in $y{P}_{\chi }$ is ${\sum }_x\chi (x)N_{y,x^{\ast }}^z={\sum }_u\chi (u^{\ast })N_{y^{\ast },z}^u$ by [F1], after setting $u=x^{\ast }$. By multiplicativity and the anti-involution identity $(y^{\ast }z{)}^{\ast }=z^{\ast }y$, this equals $\chi (z^{\ast })\chi (y)$, which is the coefficient of $z$ in $\chi (y)P_{\chi }$. Hence $y{P}_{\chi }=\chi (y)P_{\chi }$.

Applying $\psi$ to the equality in step 1.1 gives $\psi (y)\psi (P_{\chi })=\chi (y)\psi (P_{\chi })$, where $\psi (P_{\chi })={\sum }_x\chi (x)\psi (x^{\ast })$. Since $\chi \mathrm{\ne }\psi$, some $y\in A$ has $\chi (y)\mathrm{\ne }\psi (y)$; subtracting the two sides forces $\psi (P_{\chi })=0$. This is the asserted sum.

Let $K=\mathrm{\Lambda }\cap (L\oplus W)$. Projection $p_W:K\to W$ has image $R_{\mathrm{\Lambda }}(L)$ and kernel $T=\{x\in L:(x,0)\in \mathrm{\Lambda }\}$. For $(x,y),(x^{\mathrm{\prime }},y^{\mathrm{\prime }})\in K$, isotropy of $\mathrm{\Lambda }$ gives $-{\omega }_V(x,x^{\mathrm{\prime }})+{\omega }_W(y,y^{\mathrm{\prime }})=0$; isotropy of $L$ makes the first term zero. Hence $R_{\mathrm{\Lambda }}(L)$ is isotropic.

In $\bar{V}\oplus W$, the orthogonal complement of $L\oplus W$ is $L\oplus 0$, because $L^{\perp }=L$. Since $\mathrm{\Lambda }={\mathrm{\Lambda }}^{\perp }$, the orthogonal complement of $\mathrm{\Lambda }+(L\oplus W)$ is $\mathrm{\Lambda }\cap (L\oplus 0)$, of dimension $\dim T$. For any subspace $U$ of a finite-dimensional symplectic space, the map $v\mapsto \omega (v,-){\mathrm{\mid }}_U$ onto $U^{\ast }$ is surjective: extend a functional from a basis of $U$ to the ambient space and use the isomorphism defined by the nondegenerate form. Its kernel is $U^{\perp }$, so $\dim U^{\perp }=\dim V-\dim U$. Rank-nullity and [F2] therefore give $\dim K=(n+m)+(n+2m)-(2n+2m-\dim T)=m+\dim T\mathrm{.}$ The kernel of $p_W{\mathrm{\mid }}_K$ has dimension $\dim T$, so $\dim R_{\mathrm{\Lambda }}(L)=m$.

The isotropic subspace $R_{\mathrm{\Lambda }}(L)$ has dimension $m$. The formula proved in step 2.1 gives $\dim R_{\mathrm{\Lambda }}(L{)}^{\perp }=2m-m=m$. Since isotropy gives $R_{\mathrm{\Lambda }}(L)\subseteq R_{\mathrm{\Lambda }}(L{)}^{\perp }$, equality follows; this is the definition of Lagrangian in [F1].

The assignment $[a]\mapsto [b]$ is a well-defined isomorphism from $U_{123}$ to $U_{231}=\frac{L_2\cap (L_3+L_1)}{(L_2\cap L_3)+(L_2\cap L_1)}\mathrm{.}$ Indeed, changing $b$ in a fixed decomposition changes it by an element of $L_2\cap L_3$; adding an element of $L_1\cap L_2$ to $a$ changes $b$ by an element of $L_1\cap L_2$, and adding an element of $L_1\cap L_3$ can be absorbed into $c$. Conversely, for $b\in L_2\cap (L_3+L_1)$ write $b=a_0+c_0$ with $a_0\in L_1,c_0\in L_3$ and take $a=-a_0,c=-c_0$, which gives $a+b+c=0$. If $a=u_{12}+u_{13}$ with $u_{12}\in L_1\cap L_2,u_{13}\in L_1\cap L_3$, then $b=-a-c$ differs from $-u_{12}$ by an element of $L_2\cap L_3$, so its class is zero. Thus the denominator is exactly the kernel and the map is an isomorphism.

For a cyclic list $L_1,\dots ,L_n$ of Lagrangians, set $K=\ker ({\sum }_i:L_1\oplus \cdots \oplus L_n\to V)$. For $w\in K$, choose ${\widehat{w}}_i\in V$ with $w_i={\widehat{w}}_i-{\widehat{w}}_{i-1}$ cyclically; such vectors exist since ${\sum }_i{w}_i=0$, and any two choices differ by a common constant. Define $q(v,w)={\sum }_i\omega (v_i,{\widehat{w}}_i)$. Changing the common constant changes the sum by $\omega ({\sum }_i{v}_i,t)=0$. Summation by parts gives $q(v,w)={\sum }_i\omega (w_i,{\widehat{v}}_{i-1})$, which differs from $q(w,v)$ by ${\sum }_i\omega (w_i,v_i)=0$, since both entries in each pair lie in $L_i$. Thus $q$ is symmetric. Choosing ${\widehat{w}}_0=0$ and ${\widehat{w}}_i={\sum }_{j=1}^i{w}_j$ gives $q(v,w)={\sum }_i{\sum }_{j\le i}\omega (v_i,w_j)={\sum }_{i>j}\omega (v_i,w_j),$ because the diagonal terms vanish. Define $\tau (L_1,\dots ,L_n)$ as the signature of the induced nondegenerate form on $K\mathrm{/}\mathrm{rad}(q)$.

The pairing $\omega (a,b)$ descends to $U_{123}\times U_{231}$. If $a$ changes by $u_{12}+u_{13}$ with $u_{rs}\in L_r\cap L_s$, then $\omega (u_{12},b)=0$ since both lie in $L_2$, and $\omega (u_{13},b)=0$ since $b=-a-c$ and each of $u_{13},a$ lies in $L_1$ while $u_{13},c$ lie in $L_3$. Changing $b$ by $L_2\cap L_3$ also pairs to zero because $a=-b-c$. For subspaces $P,Q$, $P^{\perp }+Q^{\perp }=(P\cap Q{)}^{\perp }$: inclusion follows from the definition of orthogonality, and the dimensions agree by rank-nullity and the formula $\dim U^{\perp }=\dim V-\dim U$. The latter follows because $V\to U^{\ast }$, $v\mapsto \omega (v,-){\mathrm{\mid }}_U$, is onto (extend a functional on a basis of $U$ and use nondegeneracy of $\omega$), with kernel $U^{\perp }$. Thus $(L_2\cap (L_3+L_1){)}^{\perp }=L_2+(L_3+L_1{)}^{\perp }=L_2+(L_1\cap L_3)\mathrm{.}$ The induced pairing is nondegenerate: if $a\in L_1\cap (L_2+L_3)$ pairs to zero with $L_2\cap (L_3+L_1)$, then $a\in L_2+(L_1\cap L_3)\mathrm{.}$ Writing $a=b+t$ with $b\in L_2$ and $t\in L_1\cap L_3$, the facts $a,t\in L_1$ imply $b\in L_1\cap L_2$, so $[a]=0$. In the other variable, if $b\in L_2\cap (L_3+L_1)$ pairs to zero with every $a\in L_1\cap (L_2+L_3)$, then $b\in (L_1\cap (L_2+L_3){)}^{\perp }=L_1+(L_2\cap L_3)$. Since $b\in L_2$, its class is zero in $U_{231}$.

Fix $2\le k<n$. Let $K_A$ be the sum-zero subspace for the list $(L_1,\dots ,L_k)$ and $K_B$ that for $(L_1,L_k,L_{k+1},\dots ,L_n)$. For $a=(a_1,\dots ,a_k)$ and $b=(b_1,b_k,b_{k+1},\dots ,b_n)$ define $s(a,b)=(a_1+b_1,a_2,\dots ,a_{k-1},a_k+b_k,b_{k+1},\dots ,b_n)\in K\mathrm{.}$ Using $q(v,w)={\sum }_{i>j}\omega (v_i,w_j)$, the terms involving only $a$ give $q_A$, those involving only $b$ give $q_B$, and the cross terms reduce, by ${\sum }_i{a}_i=0=\sum b_i$, to $-\omega (a_1,b_1^{\mathrm{\prime }})+\omega (a_k^{\mathrm{\prime }},b_k)\mathrm{.}$ Both terms vanish because $a_1,b_1^{\mathrm{\prime }}\in L_1$ and $a_k^{\mathrm{\prime }},b_k\in L_k$. Thus $s$ preserves the orthogonal-sum form. Its kernel is the set of pairs with all interior coordinates zero and shared endpoint coordinates $(-u,u)$ in $K_A$ and $(u,-u)$ in $K_B$, for $u\in E:=L_1\cap L_k$. Its image is $\ker \delta$, where $\delta :K\to V\mathrm{/}(L_1+L_k),\delta (v)={\sum }_{i=1}^k{v}_i<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>(L_1+L_k)\mathrm{.}$ Indeed, the prefix sum of $s(a,b)$ is $b_1+b_k\in L_1+L_k$, because ${\sum }_i{a}_i=0$. Conversely, if the prefix $p={\sum }_{i=1}^k{v}_i$ equals $u_1+u_k$ with $u_1\in L_1,u_k\in L_k$, set $b_1=u_1,b_k=u_k$, $a_1=v_1-b_1$, $a_k=v_k-b_k$, and use $a_i=v_i$ for $1<i<k$, $b_i=v_i$ for $i>k$; then both shorter lists sum to zero and $s(a,b)=v$.

Symmetry follows from $a+b+c=a^{\mathrm{\prime }}+b^{\mathrm{\prime }}+c^{\mathrm{\prime }}=0$ and isotropy: expanding $0=\omega (a,a^{\mathrm{\prime }})=\omega (-b-c,-b^{\mathrm{\prime }}-c^{\mathrm{\prime }})$ gives $\omega (c,b^{\mathrm{\prime }})=\omega (c^{\mathrm{\prime }},b)$, and then $\omega (a,b^{\mathrm{\prime }})=\omega (a^{\mathrm{\prime }},b)$. Thus ${\psi }_{123}$ is a nondegenerate symmetric form and its signature is defined.

In finite-dimensional symplectic $V$, the map $V\to U^{\ast }$ given by $v\mapsto \omega (v,-){\mathrm{\mid }}_U$ is onto for every subspace $U$: extend a functional on $U$ to a basis of $V$ and use the isomorphism $V\to V^{\ast }$ defined by $\omega$. Its kernel is $U^{\perp }$, so $\dim U^{\perp }=\dim V-\dim U$. The inclusion $A^{\perp }+B^{\perp }\subseteq (A\cap B{)}^{\perp }$ is equality because both sides then have the same dimension. In particular $(L_1\cap L_k{)}^{\perp }=L_1+L_k$. Embed $E$ into $K$ by $r(u)=(-u,0,\dots ,0,u,0,\dots ,0)$, with entries at positions $1$ and $k$, and let $J=r(E)$. The partial-sum formula gives $q(r(u),w)=\omega \text{\hspace{0.17em}⁣}(u,{\sum }_{i=1}^k{w}_i),$ since the omitted $w_k$ term pairs to zero with $u\in L_k$. Thus $J$ is isotropic and $J^{\perp }=\ker \delta =\mathrm{im}s$, because $(L_1\cap L_k{)}^{\perp }=L_1+L_k$ by [F1]–[F2]. Put $\bar{K}=K\mathrm{/}\mathrm{rad}(q)$ and let $\bar{J}$ be the image of $J$ in $\bar{K}$. The map induced by $s$ onto ${\bar{J}}^{\perp }\mathrm{/}\bar{J}$ is surjective because $\mathrm{im}s=J^{\perp }$. Its kernel is radical by preservation of the form. Conversely, a source vector in the radical maps to a vector orthogonal to every element of ${\bar{J}}^{\perp }$; nondegeneracy of $\bar{K}$ makes that orthogonal complement $\bar{J}$, so the vector is in the kernel. Hence the nondegenerate quotient of $K_A\oplus K_B$ is isometric to ${\bar{J}}^{\perp }\mathrm{/}\bar{J}$.

For $n=3$, the partial-sum formula of step 1.2 gives $q(v,w)=\omega (v_2,w_1)+\omega (v_3,w_1)+\omega (v_3,w_2)=\omega (v_3,w_2)$, since $v_2+v_3=-v_1$ and $v_1,w_1\in L_1$. Since $v_3=-v_1-v_2$ and $v_2,w_2\in L_2$, this is $-\omega (v_1,w_2)=-{\psi }_{123}([v_1],[w_1])$. The projection $K\to U_{123}$, $(v_1,v_2,v_3)\mapsto [v_1]$, is onto. Its kernel is the image of the adjacent-intersection difference map $((u_{12},u_{23},u_{31}))⟼(u_{12}-u_{31},u_{23}-u_{12},u_{31}-u_{23})\mathrm{.}$ For the reverse inclusion, if $v_1=u_{12}+u_{13}$ represents zero, set $u_{31}=-u_{13}$ and $u_{23}=v_2+u_{12}$; then $u_{23}\in L_2\cap L_3$ by $v_1+v_2+v_3=0$, and the displayed difference is $v$. The nondegeneracy in step 2.1 now shows that this kernel is exactly $\mathrm{rad}(q)$. Consequently $\tau (L_1,L_2,L_3)=-\mu (L_1,L_2,L_3)$; this is the sign translation from the cyclic convention to Walker's Wall-form convention.

If $Q$ is a nondegenerate real symmetric form and $I$ is isotropic, choose a basis $e_1,\dots ,e_r$ of $I$ and, by nondegeneracy, vectors $f_j$ with $Q(e_i,f_j)={\delta }_{ij}$. Replace $f_j$ by $f_j-\frac{1}{2}{\sum }_{\mathrm{\ell }}Q(f_{\mathrm{\ell }},f_j)e_{\mathrm{\ell }}$; symmetry makes all pairings among the modified $f_j$ vanish, so their span with $I$ is hyperbolic of signature zero. Its orthogonal complement identifies with $I^{\perp }\mathrm{/}I$. Therefore $\mathrm{sign}(I^{\perp }\mathrm{/}I)=\mathrm{sign}(Q)$. Applying this to step 3.2 proves $\tau (L_1,\dots ,L_n)=\tau (L_1,\dots ,L_k)+\tau (L_1,L_k,\dots ,L_n)\mathrm{.}$

Reindexing a cyclic list identifies its sum-zero space with the same space and changes each chosen antiderivative only by a common constant; step 1.2 shows that this does not change $q$. Hence $\tau$ is cyclically invariant. Apply the chain identity of step 4.1 to $(L_1,L_2,L_3,L_4)$ and to its cyclic rotation $(L_2,L_3,L_4,L_1)$. Thus ${\tau }_{123}+{\tau }_{134}={\tau }_{1234}={\tau }_{234}+{\tau }_{124}\mathrm{.}$ Multiplying by $-1$ and using step 3.3 gives the asserted four-Lagrangian identity for $\mu$.

Exactness [F1] gives ${\lambda }_M=\mathrm{im}(\mathrm{\partial }:H_2(M,\mathrm{\Sigma };\mathbb{R})\to H_1(\mathrm{\Sigma };\mathbb{R}))$. By [F2], every relative class is $u\cap [M,\mathrm{\Sigma }]$ for some $u\in H^1(M;\mathbb{R})$. The cap boundary identity gives $\mathrm{\partial }(u\cap [M,\mathrm{\Sigma }])=\pm (i^{\ast }u)\cap [\mathrm{\Sigma }]$. Therefore ${\lambda }_M=P{D}_{\mathrm{\Sigma }}(\mathrm{im}{i}^{\ast })$; the sign does not change this subspace.

Let $E=\mathrm{im}{i}^{\ast }\subseteq H^1(\mathrm{\Sigma };\mathbb{R})$. For $x\in H_1(\mathrm{\Sigma };\mathbb{R})$, [F3] gives ${\omega }_{\mathrm{\Sigma }}(x,P{D}_{\mathrm{\Sigma }}\beta )=⟨\beta ,x⟩$ for $\beta \in E$. Hence $x$ is orthogonal to $P{D}_{\mathrm{\Sigma }}(E)={\lambda }_M$ exactly when every functional in $E$ vanishes on $x$.

By [F4], $E$ is the annihilator of $\ker i_{\ast }$ in $H^1(\mathrm{\Sigma };\mathbb{R})$; finite-dimensional double-annihilator duality says the vectors annihilated by every element of $E$ are exactly $\ker i_{\ast }={\lambda }_M$. Step 2.1 therefore gives ${\lambda }_M^{\perp }={\lambda }_M$. This equality implies isotropy and half-dimension, so ${\lambda }_M$ is Lagrangian. If $\mathrm{\Sigma }$ is empty, both sides are zero directly.