Braidings, cocycles and cyclic power obstructions

A tensor product can admit coherent swaps without making every double swap the identity. It can also contain an object whose power is the unit without admitting a coherent cyclic action with that object as generator. Graded vector spaces let us compute both distinctions explicitly.

We use the tensor and unit conventions of Tensor actions and absolute algebra presentations. Write \(I\) for the unit, \(a_{X,Y,Z}:(X\otimes Y)\otimes Z\to X\otimes(Y\otimes Z)\) for the associator, and \(l_X:I\otimes X\to X\), \(r_X:X\otimes I\to X\) for the unit constraints. A tensor functor has invertible tensor comparisons; a unital one also identifies the image of the unit with the target unit.

1. Coherent swaps and the unit

A braiding is a natural isomorphism \(R_{X,Y}:X\otimes Y\to Y\otimes X\) satisfying two hexagons. With composition read from right to left, their completely parenthesized expressions are

These are the standard hexagons of Etingof, Gelaki, Nikshych and Ostrik, Tensor Categories, Definition 8.1.1, printed page 195. They imply the Yang–Baxter relation of Proposition 8.1.10: the adjacent-swap sequences \(12,23,12\) and \(23,12,23\) give the same map from \(XYZ\) to \(ZYX\). Here is the full argument. The open coherence interface in Tensor actions and absolute algebra presentations, §1 lets us insert canonical associators while writing ordered words without parentheses. Naturality of \(R_{X,-}\) at \(R_{Y,Z}:YZ\to ZY\) says \[ \begin{aligned} (R_{Y,Z}\otimes1_X)R_{X,YZ} &=R_{X,ZY}(1_X\otimes R_{Y,Z}). \end{aligned} \] Expanding the two braidings with hexagon (1.1) gives \[ \begin{aligned} &(R_{Y,Z}\otimes1_X)(1_Y\otimes R_{X,Z}) (R_{X,Y}\otimes1_Z)\\ &\quad=(1_Z\otimes R_{X,Y})(R_{X,Z}\otimes1_Y) (1_X\otimes R_{Y,Z}). \end{aligned} \] Read the factors from right to left: these are exactly the two required three-swap routes. With each word left-associated, a swap at positions \(23\) is \(a^{-1}(1\otimes R)a\), with the current objects substituted in \(a\); a swap at \(12\) is \(R\otimes1\). Substitution into the expanded identity inserts precisely these reassociations, whose structural composites agree by coherence. This proves the relation in the nonstrict category as well. Author-hosted book.

Reversing the swaps gives another braiding, \[ R^-_{X,Y}=R_{Y,X}^{-1}. \] Indeed, invert (1.2) for the triple \(Y,Z,X\) to obtain (1.1) for \(R^-\); invert (1.1) for \(Z,X,Y\) to obtain its (1.2). The associators have the required reversed directions. A braiding is symmetric when \(R^-=R\), equivalently \(R_{Y,X}R_{X,Y}=1_{X\otimes Y}\). This is also called a commutative tensor category. The word “tensor” alone does not impose this condition.

Lemma 1.1. Every braiding satisfies \[ \begin{aligned} r_XR_{I,X}&=l_X,\\ l_XR_{X,I}&=r_X,\\ R_{I,I}&=1_{I\otimes I}. \end{aligned}\tag{1.3} \]

Proof. Use the coherent associativity and unit identifications to regard a word containing units as the word with those units removed. For clarity, the endomorphism induced by the first unit swap is \[ b_X=r_XR_{I,X}l_X^{-1}. \] Naturality in the first input identifies \(R_{I\otimes I,X}\), after applying the unit multiplication \(I\otimes I\to I\), with \(R_{I,X}\). Hexagon (1.2) with inputs \(I,I,X\) identifies the same map with two successive copies of \(b_X\). Hence \(b_X=b_X^2\). Since \(b_X\) is invertible, \(b_X=1_X\). Hexagon (1.1) with inputs \(X,I,I\) gives the same argument for \(l_XR_{X,I}r_X^{-1}\). Finally \(l_I=r_I\) equals the unit multiplication, so either equality at \(X=I\) gives \(R_{I,I}=1\). The unit identifications used here are precisely the coherent ones in Stacks, Lemma 4.43.4; they require no symmetry. \(\square\)

We also retain Stacks, Lemma 4.43.2: for \(\rho:I\otimes I\to I\), one has \(\rho\otimes1_I=1_I\otimes\rho\), the monoid \(\operatorname{End}(I)\) is commutative, and \[ \begin{aligned} \rho(a\otimes1_I)\rho^{-1}&=a,\\ \rho(1_I\otimes a)\rho^{-1}&=a. \end{aligned}\tag{1.4} \] The source convention that tensoring with a unit is fully faithful was upgraded to an equivalence in the preceding lesson, so the canonical unit hypothesis is available.

For any \(a:I\to I\), define natural endomorphisms of the identity by \[ \begin{aligned} L(a)_X&=l_X(a\otimes1_X)l_X^{-1},\\ R(a)_X&=r_X(1_X\otimes a)r_X^{-1}. \end{aligned}\tag{1.5} \] They need not agree in an arbitrary tensor category. In a braided category they do: naturality gives \[ R_{I,X}(a\otimes1_X) =(1_X\otimes a)R_{I,X}, \] and conjugating by (1.3) yields \(L(a)_X=R(a)_X\).

Finally, suppose \(\theta:1_{\mathcal T}\to1_{\mathcal T}\) is an invertible tensor transformation; unit compatibility has not been imposed separately. Naturality at \(\rho\), together with \(\theta_{I\otimes I}=\theta_I\otimes\theta_I\), gives \[ \theta_I=\rho(\theta_I\otimes\theta_I)\rho^{-1} =\theta_I^2 \] by (1.4). Cancellation proves \(\theta_I=1_I\). Invertibility matters for this conclusion.

2. Graded cocycles and their checked interfaces

Let \(k\) be any field and \(G\) a group, written multiplicatively in this section. Consider \(G\)-graded vector spaces of finite support. Each component may have arbitrary dimension. Morphisms preserve degrees, and \[ (V\otimes W)_t=\bigoplus_{gh=t}V_g\otimes_k W_h. \] Let \(\delta_g\) be the one-dimensional space in degree \(g\). A function \(\alpha:G^3\to k^\times\) specifies the associator by \[ (v_g\otimes w_h)\otimes z_s \longmapsto \alpha(g,h,s)\,v_g\otimes(w_h\otimes z_s). \] The canonical graded construction, Example 2.3.8 of Tensor Categories, printed page 28, makes this a tensor category exactly when \[ \begin{aligned} &\alpha(gh,s,t)\alpha(g,h,st)\\ &\quad=\alpha(g,h,s)\\ &\qquad\alpha(g,hs,t)\alpha(h,s,t). \end{aligned} \tag{2.1} \] The formula constructs a natural invertible associator: the ordinary balanced tensor relations make its homogeneous prescription well-defined, and multiplication by the inverse scalar gives its inverse. The category of all graded vector spaces contains our full subcategory, closed under tensor product because the product of two finite sets of degrees is finite. All these maps restrict to it. No finite-dimensional hypothesis has entered.

Here is the interface checking necessity and the convention for (2.1). On four homogeneous vectors, the two pentagon paths multiply by the two displayed sides. Pure tensors span every component, so equality of these scalars implies equality of the maps. Conversely, evaluate the pentagon on \(\delta_g,\delta_h,\delta_s,\delta_t\). This also checks the associator direction against the cited construction. Example 2.3.8.

No normalization assumption is needed. Put \[ p(g)=\alpha(e,e,g),\qquad q(g)=\alpha(g,e,e). \] Substituting \((e,e,g,h)\), \((g,h,e,e)\) and \((g,e,e,h)\) in (2.1), and cancelling nonzero factors, gives respectively \[ \begin{aligned} \alpha(e,g,h)&=p(gh)/p(g),\\ \alpha(g,h,e)&=q(gh)/q(h),\\ \alpha(g,e,h)&=q(g)p(h). \end{aligned} \tag{2.2} \] The all-unit substitution gives \(p(e)=q(e)=1\). Consequently \(I=\delta_e\), with left unit map on degree \(g\) equal to \(p(g)^{-1}\) times the ordinary one and right unit map equal to \(q(g)\) times it, satisfies the triangle: its two coefficients are \[ \alpha(g,e,h)p(h)^{-1}=q(g). \] At \(I\) both unit maps are the ordinary multiplication. These computations fill the unit check in the book's Exercise 2.3.9.

The cocycle change of tensor comparison is also canonical, Section 2.6 and Remark 2.6.2, printed pages 33–34. In our direction a comparison for the identity functor is \[ \xi_{V,W}:V\otimes_b W\longrightarrow V\otimes_\alpha W, \] acting on degrees \(g,h\) by a nonzero scalar \(\phi(g,h)\). It is a tensor comparison precisely when \[ \begin{aligned} &\alpha(g,h,s)\\ &\quad=\frac{\phi(h,s)\phi(g,hs)} {\phi(g,h)\phi(gh,s)}\,b(g,h,s). \end{aligned}\tag{2.3} \] To check this import, the book's map \(J: FV\otimes FW\to F(V\otimes W)\) is \(\xi^{-1}\), so its scalar is \(\phi^{-1}\). Substitution into equation (2.31) gives (2.3). Conversely, the two tensor-functor associativity paths on a homogeneous triple have coefficients \(\alpha\phi(g,h)\phi(gh,s)\) and \(b\phi(h,s)\phi(g,hs)\); their equality is (2.3). The inverse comparison uses \(\phi^{-1}\), so this identity functor is a tensor equivalence. The finite-support restriction preserves all these maps.

The book's Section 2.6 initially concerns a category with group-valued automorphisms; its linear version is the one specified in Remark 2.6.2. We use only the explicit identity comparison acting trivially on scalar maps. This respects the author's correction to Section 2.6.

One can normalize \(\alpha\) without changing its tensor equivalence class. Choose \(\phi(e,g)=p(g)^{-1}\), \(\phi(g,e)=q(g)\), and arbitrary nonzero values elsewhere, consistently with \(\phi(e,e)=1\). Apply (2.3) with \(b=\alpha\) to define the new cocycle. Equations (2.2) show that all its entries containing \(e\) are \(1\). This explicit interface is useful whenever unit maps appear in a graded calculation.

3. Scalar hexagons and three successive swaps

Now write the grading group additively as \(L\). To have swaps between every pair \(\delta_g,\delta_h\), the group must be abelian: a nonzero degree-preserving map from degree \(g+h\) to degree \(h+g\) requires equality. Assume this from now on.

For \(\rho:L^2\to k^\times\), define \[ R_{V,W}(v_g\otimes w_h)=\rho(g,h)\,w_h\otimes v_g. \] These maps are natural and invertible. Applying (1.1) and (1.2) to a homogeneous triple gives respectively \[ \begin{aligned} &\rho(g,h+s)\\ &\quad=\rho(g,h)\rho(g,s)\\ &\qquad\frac{\alpha(h,g,s)} {\alpha(g,h,s)\alpha(h,s,g)}. \end{aligned}\tag{3.1} \] \[ \begin{aligned} &\rho(g+h,s)\\ &\quad=\rho(g,s)\rho(h,s)\\ &\qquad\frac{\alpha(g,h,s)\alpha(s,g,h)} {\alpha(g,s,h)}. \end{aligned}\tag{3.2} \]

Equivalently, \[ \begin{aligned} &\frac{\alpha(g,h,s)\alpha(h,s,g)}{\alpha(h,g,s)}\\ &\quad=\frac{\rho(g,h)\rho(g,s)}{\rho(g,h+s)}\\ &\quad=\frac{\rho(h+s,g)}{\rho(h,g)\rho(s,g)}. \end{aligned} \tag{3.3} \] For example, (1.1) has five factors, whose scalars are \(\alpha(g,h,s)^{-1}\), \(\rho(g,h)\), \(\alpha(h,g,s)\), \(\rho(g,s)\), and \(\alpha(h,s,g)^{-1}\). This gives (3.1) with every inverse accounted for. Reading (1.2) in the same order gives (3.2).

Pure tensors prove sufficiency; the three graded lines prove necessity. Relabel \((g,h,s)\) in (3.2) as \((h,s,g)\) to combine the two hexagons into (3.3). Thus apparently different scalar forms of the second hexagon agree after a cyclic relabeling.

There is a weaker test. Start and end with left-associated triples. Put \(P=\rho(g,h)\rho(g,s)\rho(h,s)\). The swap sequence \(12,23,12\) has scalar \[ P\,\frac{\alpha(h,g,s)}{\alpha(h,s,g)}. \] The sequence \(23,12,23\) has scalar \[ P\,\frac{\alpha(g,h,s)\alpha(s,g,h)} {\alpha(g,s,h)\alpha(s,h,g)}. \] The two maps agree exactly when \[ \begin{aligned} &\alpha(g,h,s)\alpha(h,s,g)\\ &\quad\alpha(s,g,h)\\ &\qquad=\alpha(g,s,h)\alpha(s,h,g)\\ &\qquad\quad\alpha(h,g,s). \end{aligned} \tag{3.4} \] Here the nonzero factor \(P\) cancels. This is the scalar Yang–Baxter test; it need not imply (3.3).

The symmetric condition has a separate, simpler form: \[ \rho(g,h)\rho(h,g)=1. \tag{3.5} \] It follows by computing the double swap on homogeneous vectors, with no associator involved.

4. A twisted parity example and tensor endomorphisms

Take \(L=\mathbb Z/2\mathbb Z\), represented by bits \(0,1\), and put \[ \alpha(g,h,s)=(-1)^{ghs}. \tag{4.1} \] To verify the cocycle condition, compute exponents modulo \(2\). The exponent on its left is \((g+h)st+gh(s+t)\); on its right it is \(ghs+g(h+s)t+hst\). Both expand to \(ghs+ght+gst+hst\). Thus (4.1) is a cocycle over every field, including characteristic \(2\), where it is trivial.

It is normalized, so Lemma 1.1 requires \[ \rho(0,0)=\rho(0,1)=\rho(1,0)=1. \] Write \(u=\rho(1,1)\). In (3.3) the triple \((1,1,1)\) gives \(u^2=-1\). If a triple contains a zero, the associator ratio is \(1\); for \(g=0\) both ratios of \(\rho\) are \(1\), and for \(h=0\) or \(s=0\) numerator and denominator cancel. These are all remaining triples. Consequently the braidings are exactly \[ \rho(1,1)=u,\qquad u^2=-1, \tag{4.2} \] with all other entries \(1\). If \(i\in k\) satisfies \(i^2=-1\), the possibilities are \(u=i,-i\); they coincide in characteristic \(2\). In characteristic different from \(2\), the double swap of two odd lines is \(-1\), so this braiding is not symmetric.

For the trivial associator, \(\rho(g,h)=(-1)^{gh}\) instead satisfies both hexagons: each variable is additive modulo \(2\) in the exponent. Its double swap is \(1\), giving the usual symmetric super vector spaces. This is the parity specialization of the canonical Example 8.2.2, printed page 197; the homogeneous checks extend it to unrestricted component dimensions.

For any of these graded tensor categories, a function \(\psi:L\to k\) defines a natural transformation of the identity by scalar multiplication \(\psi(g)\) on degree \(g\). Binary tensor compatibility holds exactly when \[ \psi(g+h)=\psi(g)\psi(h). \tag{4.3} \] Necessity follows on graded lines, and sufficiency on pure tensors. A tensor transformation in this binary sense need not be invertible; in particular the zero function is allowed when unit compatibility is not required. For a group, (4.3) implies either that \(\psi\) is identically zero or that \(\psi(0)=1\) and \(\psi(g)\psi(-g)=1\). In the latter case it is a character into \(k^\times\), and the transformation is invertible. Its unit component is then the identity, as the general argument in Section 1 predicts.

5. When a power relation gives a cyclic tensor functor

Let \(\mathcal T\) be any tensor category with unit, let \(n>0\), and choose an isomorphism \(\lambda:X^{\otimes n}\to I\). Normalize every word to left-associated parentheses using the canonical coherence maps. On \(X^{\otimes(n+1)}\) there are two contractions: \[ \begin{aligned} d_L&=l_X(\lambda\otimes1_X),\\ d_R&=r_X(1_X\otimes\lambda)\,A. \end{aligned} \tag{5.1} \] Here \(A:X^{\otimes(n+1)}\to X\otimes X^{\otimes n}\) is that canonical reassociation. Suppose \(d_L=d_R\).

Proposition 5.1. There is a unital tensor functor from the discrete tensor category \(\mathbb Z/n\mathbb Z\) to \(\mathcal T\), taking the residue of \(1\) to \(X\).

Proof. First take \(n\ge2\), and represent residues by \(0,\ldots,n-1\). Set \(F(r)=X^{\otimes r}\), with \(F(0)=I\). To multiply \(F(r)\otimes F(s)\), concatenate the words. If \(r+s<n\), only reassociate. Otherwise apply \(\lambda\) to a consecutive block of \(n\) copies and remove the resulting unit. This defines an isomorphism \[ M_{r,s}:F(r)\otimes F(s)\longrightarrow F(r+s\bmod n). \]

We verify that the choice of block and all triple products are coherent. In a word of length at least \(n+1\), sliding a contracted block one position replaces \(d_L\) by \(d_R\), tensored with the unchanged prefix and suffix. Equality in (5.1) proves that slide changes no map. Repeated slides show that any single block contraction gives the same normalized map \(c_m:X^{\otimes m}\to X^{\otimes(m-n)}\). After one contraction the remaining word again consists entirely of copies of \(X\); thus any \(q\) successive contractions give \[ c_{m-(q-1)n}\cdots c_{m-n}c_m. \] This proves independence for multiple contractions as well.

For a triple of residues \(r,s,t\), either parenthesized product removes exactly \(\lfloor(r+s+t)/n\rfloor\) blocks. The preceding argument identifies both resulting maps from the same concatenated word, and canonical reassociation identifies their sources. This is the tensor-functor associativity equation. Products involving \(0\) use \(l,r\); their equations are the unit coherence identities. The maps \(M_{r,s}^{-1}\) are the tensor comparisons in the direction used in Section 2. Since the source is discrete, there are no further morphism naturality conditions.

If \(n=1\), the source has one object, and the residue of \(1\) is \(0\). Set \(F(0)=X\), identify its unit with \(I\) by \(\lambda\), and transport the unit multiplication to \[ X\otimes X\xrightarrow{\lambda\otimes\lambda} I\otimes I\xrightarrow{\rho}I\xrightarrow{\lambda^{-1}}X. \] Its associativity and unit laws are transported from those of \(I\). This deals with the positive-integer edge case without requiring \(F(0)\) literally to equal \(I\). \(\square\)

Proposition 5.2. In a braided category, the equality \(d_L=d_R\) is independent of the chosen isomorphism \(\lambda\).

Proof. Another isomorphism is \(\lambda'=a\lambda\) for a unique \(a\in\operatorname{Aut}(I)\). The associated maps satisfy \(d_L'=L(a)_Xd_L\) and \(d_R'=R(a)_Xd_R\). Section 1 gives \(L(a)_X=R(a)_X\); this common map is invertible. Therefore the original two contractions agree if and only if the new two do. \(\square\)

Existence of a braiding does not force the contractions to agree. In the twisted parity category, take \(X=\delta_1\), \(n=2\), and any nonzero scalar \(\lambda:X\otimes X\to I\). On the triple of odd lines, \(d_L\) has coefficient \(\lambda\), while \(d_R\) has coefficient \(-\lambda\) because \(A=a_{X,X,X}\) has coefficient \(-1\). When \(\operatorname{char}k\ne2\) they differ. Over \(k=\mathbb C\), (4.2) provides braidings, so this is also a braided counterexample.

6. An invariant which detects the twisted category

The twisted and untwisted parity categories have the same underlying category. Nevertheless, in characteristic different from \(2\) they are not equivalent as tensor categories, even if an equivalence is not assumed to be \(k\)-linear.

We prove this using (5.1). First, a tensor-invertible object in the untwisted graded category is one-dimensional and supported in one degree. Indeed, if \(V\otimes W\cong I\), both total vector spaces are nonzero and their tensor product has dimension one. Choose \(w\ne0\) and a linear functional \(f\) with \(f(w)=1\). The map \(v\mapsto v\otimes w\) is split injective by \(1\otimes f\), so \(V\) has dimension at most one; reversing the roles gives the same for \(W\). The grading of a one-dimensional space has exactly one nonzero component.

Every one-dimensional graded \(Y\) with \(Y^2\cong I\) in the untwisted category satisfies \(d_L=d_R\): choose a basis \(y\), write \(\lambda(y\otimes y)=c\), and both contractions send \(y\otimes y\otimes y\) to \(cy\). This covers both possible degrees and every isomorphism \(\lambda\).

In the twisted category \(X=\delta_1\) is tensor-invertible, because \(X\otimes X\cong I\), but its contractions disagree. A unital strong tensor equivalence preserves tensor-invertibility and transports the diagram (5.1), including its associator and unit maps. It is faithful, so it also reflects equality of its two arrows. Its image would be a tensor-invertible \(Y\) in the untwisted category with a transported power isomorphism, contradicting the preceding computation. A strong tensor equivalence not initially specified as unital is unital under the source convention: essential surjectivity transfers the fully faithful unit tensor functors, and the image unit multiplication is the transported one. Thus the obstruction excludes that case too.

This argument neither assumes that the equivalence fixes the graded lines nor relies on a classification of group cohomology.

7. Exercises with complete solutions

Exercise 1 (basic). Let \(L=\mathbb Z/n\mathbb Z\), \(n>0\), and let \(\psi:L\to k\) satisfy (4.3). Classify the resulting binary tensor endomorphisms of the identity. Which are invertible, and which respect the unit?

Solution. If \(\psi(0)=0\), then \(\psi(g)=\psi(g+0)=\psi(g)\psi(0)=0\) for every \(g\). Otherwise \(\psi(0)^2=\psi(0)\) implies \(\psi(0)=1\); the relation \(\psi(g)\psi(-g)=1\) makes every value nonzero. Put \(\zeta=\psi(1)\). Repeated use of (4.3) gives \(\psi(r)=\zeta^r\), and \(\zeta^n=\psi(0)=1\). Conversely every \(\zeta\in k^\times\) with \(\zeta^n=1\) gives a well-defined such function on residues, since changing a representative by \(n\) leaves its value unchanged. These and the zero function exhaust the possibilities. The characters give invertible transformations, with inverse scalars \(\zeta^{-r}\), and have unit value \(1\). The zero transformation is neither invertible nor unit-preserving, since \(I=k\ne0\). For \(n=1\), the only character is the constant function \(1\).

Exercise 2 (intermediate). In the cyclic graded category with trivial associator, classify all scalar braidings. Determine when they are symmetric, and whether the line \(\delta_1\) admits a cyclic tensor realization.

Solution. Equations (3.1) and (3.2) say that \(\rho\) is multiplicative in each variable. The unit values are \(1\). Put \(q=\rho(1,1)\). Successive additions show \(\rho(a,b)=q^{ab}\), and periodicity gives \(q^n=1\). Conversely this formula is independent of both representatives when \(q^n=1\), and it is multiplicative in each variable, so it defines a braiding. Its double swap is \(q^{2ab}\); it is symmetric exactly when \(q^2=1\), detected at \(a=b=1\). Choose the ordinary basis identification \(\delta_1^{\otimes n}\cong\delta_0\). Both contractions in (5.1) send the tensor of \(n+1\) basis elements to the same basis element. Proposition 5.1 therefore supplies the cyclic tensor functor, for every braiding just classified. No preservation of that braiding by the functor was requested. For \(n=1\), all values and the realization are the unit case.

Exercise 3 (advanced). Assume \(\operatorname{char}k\ne2\). In the twisted parity category put \(\rho(g,h)=1\) for every pair. Prove that its swaps satisfy Yang–Baxter but do not form a braiding. Explain exactly what changes if \(k\) contains \(i\) with \(i^2=-1\).

Solution. The cocycle \((-1)^{ghs}\) is unchanged by every permutation of its inputs. Therefore the three factors on each side of (3.4) coincide, and the parenthesized Yang–Baxter maps agree for these swaps. For the triple \((1,1,1)\), the left side of (3.3) is \((-1)(-1)/(-1)=-1\), whereas both ratios formed from \(\rho=1\) are \(1\). Since \(-1\ne1\), the hexagons fail. If the field contains \(i\), replace the odd-odd value by either \(i\) or \(-i\), retaining the other values \(1\). Section 4 checks every triple and proves that these, and only these, are braidings. Their odd double swap is \(i^2=-1\), so they still are not symmetric. Their existence also leaves the cyclic contraction obstruction unchanged, since that obstruction involves the associator and \(\lambda\), rather than the choice of swaps.

Exercise 4 (expert). Work with \(\mathbb C\)-bimodules whose left and right \(\mathbb R\)-actions agree, and tensor over \(\mathbb C\). The unit is the ordinary bimodule \(\mathbb C\). Let \(X\) have underlying space \(\mathbb C\), usual left action, and right action \(x\cdot z=x\overline z\). Show that \(\lambda(x\otimes y)=x\overline y\) is a bimodule isomorphism \(X\otimes_{\mathbb C}X\to I\). Compare the contractions for \(\lambda\) and for \(i\lambda\). Deduce why the braided hypothesis in Proposition 5.2 matters.

Solution. This tensor category is the bimodule construction of Example 2.3.13, printed page 29 of Tensor Categories, applied to the \(\mathbb R\)-algebra \(\mathbb C\). The indicated subcategory uses the same real action on both sides and is closed under this tensor product.

The balancing relation is \(x\overline a\otimes y=x\otimes ay\). Both sides are sent by \(\lambda\) to \(x\overline a\,\overline y\). Left multiplication is plainly respected; the right action gives \[ \lambda(x\otimes y\overline a)=x\overline y\,a, \] as required for the ordinary right action on \(I\). The inverse is \(z\mapsto z\otimes1\): balancing gives \(x\otimes y=(x\overline y)\otimes1\), and the other composite is immediately the identity.

For a triple \(x\otimes y\otimes z\), the left contraction is the left action of \(x\overline y\) on \(z\), namely \(x\overline y z\). The right contraction is the right action of \(y\overline z\) on \(x\), namely \(x\overline{y\overline z}=x\overline y z\). Thus \(\lambda\) satisfies (5.1). Replacing it by \(i\lambda\) multiplies the left result by \(i\), but the right result by \(\overline i=-i\). They differ already at \(x=y=z=1\). The choice of power isomorphism therefore matters in this tensor category. In fact it cannot admit a braiding: for \(a=i\in\operatorname{End}(I)\), formula (1.5) gives \(L(i)_X(x)=ix\) and \(R(i)_X(x)=-ix\), contradicting the equality forced by a braiding.

8. References and status

The canonical constructions and imports above are Etingof, Gelaki, Nikshych and Ostrik, Tensor Categories, Examples 2.3.8 and 2.3.13, Section 2.6 with Remark 2.6.2, Definition 8.1.1, Proposition 8.1.10 and Example 8.2.2; the relevant omitted unit checks are written out here. Read the book through its author-hosted final version, together with the author's corrections. Its text is cited rather than included in this course.

The unit interfaces are Stacks, Lemma 4.43.2 and Lemma 4.43.4. Monoidal and braided categories are treated in Pavel Etingof, Shlomo Gelaki, Dmitri Nikshych and Victor Ostrik, Tensor Categories, author's final version, Chapters 2 and 8. The cocycle, gauge, hexagon and power-contraction formulas above give the constructions and their proofs.

Self-checked; no independent review. Written by GPT-6.1 Sol at Ultra reasoning effort. Original exposition and solutions are dedicated under CC0 1.0; cited books and the Stacks Project retain their own licenses. This lesson is a draft within a course whose full source and dependency reconciliation remains in progress.