Modular Categories, Verlinde Formula, and TQFT: Examples and Counterexamples

The projectors $(1+T)\mathrm{/}2$ and $(1-T)\mathrm{/}2$ split every representation as $V=V_{+}\oplus V_{-}$, with $T=+1$ on $V_{+}$ and $T=-1$ on $V_{-}$. Choosing finite bases decomposes these spaces into copies of the one-dimensional trivial representation $1$ and sign representation $\epsilon$. They are nonisomorphic simples, their endomorphism rings are $\mathbb{C}$, and $\epsilon \otimes \epsilon \cong 1$ since $(-1{)}^2=1$. Thus the category is finite semisimple with simple unit; ordinary duality and flip make it symmetric ribbon with the stated spherical structure. Contraction of the ordinary evaluation and coevaluation is the ordinary linear trace, so both simple dimensions are one. By [F2], for any simple labels $i,j$, $s_{ij}=\mathrm{tr}(1_{i\otimes j})=d_i{d}_j=1$. In the ordered basis $(1,\epsilon )$ this gives $s=\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)\mathrm{.}$

The two rows coincide, and $\det (s)=1\cdot 1-1\cdot 1=0$. Therefore $s$ is not invertible and the category is not modular. This shows why nondegeneracy must be a hypothesis or separately proved.

Normalization holds because every exponent vanishes if any argument is $0$. For bits, $b+c$ in the cocycle equations is reduced modulo $2$. The pentagon signs have exponents $abc+a(b+c)d+bcd=acd+bcd+abc+abd(\mathrm{m}\mathrm{o}\mathrm{d}2),$ so its two sides agree. For the first hexagon, the ratio of the $r$-terms is $i^{-2abc}=(-1{)}^{abc}$, which is exactly the ratio of the $\omega$-terms. For the second hexagon, the ratio of the $r$-terms is again $i^{-2abc}=(-1{)}^{abc}$, equal to $\omega (a,b,c)\omega (c,a,b)\omega (a,c,b{)}^{-1}=(-1{)}^{abc}$. The twist equation is trivial if either bit is zero; at $(1,1)$ its right side is $r(1,1{)}^{2}q(1{)}^2=i^2{i}^2=1=q(0)$. Also $q(-g)=q(g)$. Thus all specified cocycle and twist identities hold.

Supply the dual and pivotal data required by [F1]. The left dual of $g$ is $g$; its evaluation scalar is $1$, and its coevaluation scalar is $\omega (g,g,g{)}^{-1}=(-1{)}^g$. Both snake composites are one, since the associator contributes the same sign. Formula [F3], with reassociation from $g\otimes (g\otimes g)$ to $(g\otimes g)\otimes g$, gives $u_0=1$ and $u_1=(-1)(-1)i=i$. Put $j_g=u_{g}q(g)$, so $j_0=1$ and $j_1=-1$, and extend by direct sums. The tensor identity in [F3] and the balancing identity checked in step 1.1 prove monoidality of $j$, including all associators and double-dual comparisons. Its pivotal trace on a simple is the scalar ${\mathrm{ev}}_{g^{\vee }}(j_g\otimes 1){\mathrm{coev}}_g=(-1{)}^g(-1{)}^g=1$. The dual grade is the same, so the left and right simple traces agree; additivity gives sphericality on all finite direct sums. The induced twist $u^{-1}j$ is exactly $q$. Thus the full dimension-one pointed ribbon input, rather than just its scalar cocycle, has been constructed.

The double-braiding values are $b(0,g)=b(g,0)=1$ and $b(1,1)=i^2=-1$. Hence $s=\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)\mathrm{.}$ The only nonzero element $1$ is not in the radical because $b(1,1)=-1\mathrm{\ne }1$, so $\mathrm{rad}(b)=\{0\}$. By [F2] the category is modular; directly, $\det (s)=-2$ and $D^2=2$.

The labels are $0,1$. The dimension formula gives $d_0=d_1=1$. The twist formula gives ${\theta }_0=1$ and ${\theta }_1=q^{3\mathrm{/}2}=e^{\pi i\mathrm{/}2}=i$. Thus $D^2=1+1=2$ and choose $D=\sqrt{2}$. The Gauss sums are $p_{+}=1+i,p_{-}=1-i,\kappa =\frac{1+i}{\sqrt{2}}=e^{\pi i\mathrm{/}4}\mathrm{.}$ In particular $\kappa \mathrm{\ne }1$.

A positive blow-up multiplies the surgery link value by $p_{+}$ and changes $(m,\sigma )$ to $(m+1,\sigma +1)$, so its normalized ratio is $D^{-1}{\kappa }^{-1}{p}_{+}=D^{-1}\frac{D}{p_{+}}{p}_{+}=1.$ A negative blow-up has ratio $D^{-1}\kappa p_{-}=D^{-1}\frac{p_{+}{p}_{-}}{D}=1$ by [F2]. If the factor ${\kappa }^{-\sigma }$ is omitted, the same two blow-ups multiply the value by $p_{+}\mathrm{/}D=\kappa$ and $p_{-}\mathrm{/}D={\kappa }^{-1}$, both nontrivial. Thus an unweighted surgery link sum is not invariant under these moves for this actual category; this is a failure of omitting the signature correction, not an assertion that the weight-zero arrows form a subcategory.

The actual modular quotient is supplied by [A1], so the preceding values are realized category data. The full weighted functor uses ${\kappa }^{n+a-\sigma (B)}$ together with its exact outgoing cap factors. A Kirby blow-up changes neither the marked bordism, its weight nor its cap-reference correction, and step 2.1 cancels its complete surgery ratio. A unit increase in bordism weight instead multiplies the map by $\kappa =e^{\pi i\mathrm{/}4}\mathrm{\ne }1$. Thus the actual category has a nontrivial framing anomaly, and the proved weighted functor includes it with the positive weight convention.

At genus $g=2$ and with no marked labels, [F2] gives $\dim V({\mathrm{\Sigma }}_2)={\sum }_{x=0}^2{S}_{0x}^{-2}\mathrm{.}$

The three terms are $(1\mathrm{/}2{)}^{-2}=4$, $(1\mathrm{/}\sqrt{2}{)}^{-2}=2$, and $(1\mathrm{/}2{)}^{-2}=4$. Their sum is $10$.

Evaluate the sine dimension formula at $a=0,1,2$: it gives $1,\sqrt{2},1$. The twist exponents $a(a+2)\mathrm{/}2$ are $0,3\mathrm{/}2,4$, so $\theta =(1,e^{3\pi i\mathrm{/}8},e^{\pi i})=(1,e^{3\pi i\mathrm{/}8},-1)$. Substitution into the sine matrix formula gives the displayed entries.

The truncated fusion rule gives $1\otimes 1$ outputs $0,2$, $1\otimes 2$ output $1$, and $2\otimes 2$ output $0$, each with multiplicity one. Since there are three simple labels, [F3] gives torus dimension $3$.

The dimension formula gives $d_0=d_1=1$. The twist exponents are $0$ and $3\mathrm{/}2$, so ${\theta }_0=1$ and ${\theta }_1=e^{\pi i\mathrm{/}2}=i$. Substitution in the sine formula gives $S_{00}=S_{01}=S_{10}=1\mathrm{/}\sqrt{2}$ and $S_{11}=-1\mathrm{/}\sqrt{2}$. The truncated fusion range for $1\otimes 1$ contains only $0$.

There are exactly two labels; hence the torus dimension formula gives $\dim V(T^2)=2$. Also $S^2=I$ by direct multiplication of the displayed matrix, verifying nondegeneracy in this finite case.