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The Drinfeld–Jimbo formulas define a Hopf algebra

Statement

Let $T$ and $I_{D J}$ be the free $Q (q)$ -algebra and defining ideal of The Drinfeld-Jimbo quantized enveloping algebra by generators and relations, and write $K_{i} = K_{d_{i} h_{i}}$ . Let $Δ : T \to T \otimes_{Q (q)} T$ and $ε : T \to Q (q)$ be the algebra homomorphisms determined by

$Δ (K_{h}) = K_{h} \otimes K_{h}, Δ (E_{i}) = E_{i} \otimes K_{i}^{- 1} + 1 \otimes E_{i}, Δ (F_{i}) = F_{i} \otimes 1 + K_{i} \otimes F_{i}, ε (K_{h}) = 1, ε (E_{i}) = ε (F_{i}) = 0.$

Let $S : T \to T$ be the $Q (q)$ -linear algebra anti-homomorphism determined by

$S (K_{h}) = K_{- h}, S (E_{i}) = - E_{i} K_{i}, S (F_{i}) = - K_{i}^{- 1} F_{i} .$

Then

$Δ (I_{D J}) \subseteq I_{D J} \otimes T + T \otimes I_{D J}, ε (I_{D J}) = 0, S (I_{D J}) \subseteq I_{D J} .$

Hence these maps descend to $U_{q} (g) = T / I_{D J}$ and make it a Hopf algebra over $Q (q)$ (Bialgebras, counits and antipodes over a commutative ring): $Δ$ is coassociative, $ε$ is a counit, and

$m (S \otimes id) Δ = η \circ ε = m (id \otimes S) Δ .$

The antipode $S$ is unique for this multiplication, unit, coproduct and counit (Uniqueness of the antipode), and its square satisfies

$S^{2} (E_{i}) = q_{i}^{- 2} E_{i}, S^{2} (F_{i}) = q_{i}^{2} F_{i}, S^{2} (K_{h}) = K_{h} .$

Facts & Assumptions

Given: A symmetrizable Cartan datum over $Q (q)$ and its Drinfeld–Jimbo presentation.

[F1]

The symmetrizer satisfies $d_{i} a_{i j} = d_{j} a_{j i}$ , so the two crossing factors $q_{i}^{a_{i j}}$ and $q_{j}^{a_{j i}}$ are equal (Symmetrizable Cartan data for quantum groups).

[F2]

The defining ideal is generated by the toral, toral-action, mixed $E_{i} F_{j}$ , and positive and negative Serre relation families (The Drinfeld-Jimbo quantized enveloping algebra by generators and relations).

[F3]

The $q_{i}$ -Gaussian coefficients are symmetric in $r$ and $m - r$ , and $q_{i} - q_{i}^{- 1} \neq 0$ (Quantum integers, factorials, Gaussian binomials and divided powers at $q_{i}$ ).

[F4]

In the toral-action algebras, the Serre coproducts are the quasiprimitive formulas stated in The coproduct preserves the positive and negative quantum Serre ideals; their images in $U_{q} (g) \otimes U_{q} (g)$ are zero.

[F5]

The tensor product of algebras has multiplication $(a \otimes b) (c \otimes d) = a c \otimes b d$ (The tensor product of $R$ -algebras has multiplication $(a \otimes b) (a^{'} \otimes b^{'}) = a a^{'} \otimes b b^{'}$ ).

[F6]

Generator assignments extend uniquely to algebra maps on the free tensor algebra (Universal property of the tensor algebra).

[F7]

A map that kills a two-sided ideal factors through the quotient ring (The quotient ring $R / I$ with $(r + I) (s + I) = r s + I$ , A ring homomorphism whose kernel contains a two-sided ideal factors uniquely through the quotient ring).

[F8]

A two-sided ideal generated by a subset is as in The ideal generated by a subset and principal ideals.

[F9]

Tensoring an exact sequence ending in zero preserves exactness at the two rightmost terms (Tensoring is right exact).

[F10]

Tensor products over $Q (q)$ have the symmetry isomorphism $x \otimes y \mapsto y \otimes x$ (Symmetry and associativity isomorphisms for tensor products over a commutative ring).

[F11]

A bialgebra has algebra maps $Δ, ε$ satisfying coassociativity and the counit equations; a Hopf algebra has an antipode satisfying both convolution-inverse equations (Bialgebras, counits and antipodes over a commutative ring).

[F12]

An antipode, if it exists, is unique (Uniqueness of the antipode).

Proof

technique · Check the defining relations after mapping into $U_q(\mathfrak g)\otimes U_q(\mathfrak g)$, use tensor right exactness to recover the free-ideal coideal inclusion, and verify the Hopf equations on generators with an explicit word-induction step for convolution

1.1givenF5F6construct

The displayed assignments extend from the free generators to an algebra map $Δ : T \to T \otimes T$ by [F6], using the tensor product algebra structure in [F5]. The counit assignment extends to an algebra map by [F6], and reversing words extends the assignment for $S$ to a $Q (q)$ -linear anti-homomorphism.

1.2givenF7F9F10algebra

Let $π : T \to U : = T / I_{D J}$ be the quotient map. Write $I \otimes T$ and $T \otimes I$ for the images of the corresponding tensor-product maps in $T \otimes T$ . Right exactness [F9] gives $\ker (π \otimes {id}_{T}) = im (I \otimes T \to T \otimes T)$ ; by applying [F9] again after the flip [F10], $\ker ({id}_{U} \otimes π) = im (U \otimes I \to U \otimes T)$ , and $π \otimes {id}_{I} : T \otimes I \to U \otimes I$ is surjective. If $z \in \ker (π \otimes π)$ , then $(π \otimes {id}_{T}) z$ lies in this second kernel, so choose $y \in T \otimes I$ with the same image; then $z - y \in \ker (π \otimes {id}_{T})$ . Thus $\ker (π \otimes π) = I \otimes T + T \otimes I$ . This uses only a lift for one tensor at a time, not a choice function.

1.3givenF2F3algebra

The counit sends $K_{0} - 1$ and $K_{h} K_{h^{'}} - K_{h + h^{'}}$ to $1 - 1 = 0$ . It sends each toral-action relation to $0 - 0$ , each mixed relation to $0 - 0 - δ_{i j} (1 - 1) / (q_{i} - q_{i}^{- 1}) = 0$ , and every Serre sum to zero because each monomial contains an $E$ or an $F$ . Thus $ε (I_{D J}) = 0$ .

1.4givenF2F7algebra

The anti-homomorphism $S$ sends $K_{0} - 1$ to itself and $K_{h} K_{h^{'}} - K_{h + h^{'}}$ to $K_{- h^{'}} K_{- h} - K_{- (h + h^{'})}$ , another toral generator. For $b = ⟨ α_{i}, h ⟩$ , applying $S$ to the $E_{i}$ toral-action relation gives $- E_{i} K_{i} K_{- h} + q^{b} K_{- h} E_{i} K_{i} = 0$ in $U$ by the relation at $- h$ ; applying it to the $F_{i}$ relation gives $- K_{i}^{- 1} F_{i} K_{- h} + q^{- b} K_{- h} K_{i}^{- 1} F_{i} = 0$ by the same relation and commutation of the toral elements.

2.1step 1.1F2F5algebra

In $U \otimes U$ , the toral relations are preserved: $Δ (K_{0}) = 1 \otimes 1$ and $Δ (K_{h} K_{h^{'}}) = K_{h + h^{'}} \otimes K_{h + h^{'}}$ . For $b = ⟨ α_{i}, h ⟩$ , the two terms in $Δ (K_{h}) Δ (E_{i}) - q^{b} Δ (E_{i}) Δ (K_{h})$ cancel by $K_{h} E_{i} = q^{b} E_{i} K_{h}$ in each tensor factor and commutation of toral elements; the two terms in $Δ (K_{h}) Δ (F_{i}) - q^{- b} Δ (F_{i}) Δ (K_{h})$ cancel by the negative toral-action relation in each factor. Therefore $(π \otimes π) Δ$ kills the toral and toral-action generators of $I_{D J}$ .

2.2step 1.4F1F2F3F5algebra

Applying $S$ to $R_{i j} = E_{i} F_{j} - F_{j} E_{i} - δ_{i j} C_{i}$ and moving toral factors with the defining relations gives $S (R_{i j}) = q_{j}^{2} q_{i}^{- a_{i j}} (F_{j} E_{i} - E_{i} F_{j}) K_{i} K_{j}^{- 1} - δ_{i j} S (C_{i})$ in $U$ . For $i \neq j$ the first factor vanishes by $E_{i} F_{j} = F_{j} E_{i}$ ; for $i = j$ its scalar is $q_{i}^{2} q_{i}^{- 2} = 1$ , $F_{i} E_{i} - E_{i} F_{i} = - C_{i}$ , and $S (C_{i}) = - C_{i}$ , so the two terms cancel. Hence $S (R_{i j}) = 0$ in $U$ .

2.3step 1.4F1F2F3algebra

Let $m = 1 - a_{i j}$ . In $S ({S e r r e}_{i j}^{+})$ , each of the $m + 1$ substituted generators contributes a minus sign; moving the inserted $K_{i}, K_{j}$ to the right contributes exponent $s (s - 1) + (m - s) (m - s - 1) + a_{i j} m + 2 s (m - s) = m (m - 1 + a_{i j}) = 0$ . Reindexing $s \mapsto m - s$ and using the symmetric Gaussian coefficients [F3] contributes the remaining sign, so $S ({S e r r e}_{i j}^{+}) = - {S e r r e}_{i j}^{+} K_{i}^{m} K_{j}$ . For the negative family, $K_{i}^{- 1} F_{i} = q_{i}^{2} F_{i} K_{i}^{- 1}$ gives $(K_{i}^{- 1} F_{i})^{r} = q_{i}^{r (r + 1)} F_{i}^{r} K_{i}^{- r}$ by induction. Moving all toral factors to the right contributes exponent $s (s + 1) + (m - s) (m - s + 1) + 2 s (m - s) + a_{i j} m = m (m + 1 + a_{i j}) = 2 m$ ; the equality $q_{j}^{a_{j i}} = q_{i}^{a_{i j}}$ from [F1] is used when $K_{j}^{- 1}$ crosses $F_{i}$ , and $K_{j}^{- 1} F_{j} = q_{j}^{2} F_{j} K_{j}^{- 1}$ supplies the additional factor $q_{j}^{2}$ . Reindexing $s \mapsto m - s$ and using [F3] gives $S ({S e r r e}_{i j}^{-}) = - q_{i}^{2 m} q_{j}^{2} {S e r r e}_{i j}^{-} K_{i}^{- m} K_{j}^{- 1}$ . These identities are computed modulo the toral and toral-action relations only; their right sides lie in the two-sided Serre ideal. Thus both Serre families map into $I_{D J}$ .

3.1step 1.1step 2.1F1F2F3F5algebra

Put $a = a_{i j}$ and $C_{i} = (K_{i} - K_{i}^{- 1}) / (q_{i} - q_{i}^{- 1})$ . Expanding the commutator of the coproducts, the cross terms cancel because $E_{i} K_{j} = q_{i}^{- a} K_{j} E_{i}$ and $K_{i}^{- 1} F_{j} = q_{i}^{a} F_{j} K_{i}^{- 1}$ , where the equality $q_{i}^{a_{i j}} = q_{j}^{a_{j i}}$ is [F1]. Hence $[Δ (E_{i}), Δ (F_{j})] = (E_{i} F_{j} - F_{j} E_{i}) \otimes K_{i}^{- 1} + K_{j} \otimes (E_{i} F_{j} - F_{j} E_{i})$ . In $U$ , $E_{i} F_{j} - F_{j} E_{i} = δ_{i j} C_{i}$ , and direct expansion gives $Δ (C_{i}) = (K_{i} \otimes K_{i} - K_{i}^{- 1} \otimes K_{i}^{- 1}) / (q_{i} - q_{i}^{- 1}) = C_{i} \otimes K_{i}^{- 1} + K_{i} \otimes C_{i}$ . Thus the image under $(π \otimes π) Δ$ of every mixed relation is zero.

3.2step 2.1F4algebra

For each positive or negative Serre generator, [F4] maps its coproduct to zero in $U \otimes U$ . Therefore $(π \otimes π) Δ$ kills all remaining generators of $I_{D J}$ .

3.3step 1.4step 2.2step 2.3F2F7F8algebra

Steps 1.4, 2.2 and 2.3 show that $S$ sends every generator of $I_{D J}$ into $I_{D J}$ . Since $S$ reverses products and $I_{D J}$ is a two-sided generated ideal, $S (I_{D J}) \subseteq I_{D J}$ ; the quotient property [F7] gives the descended anti-homomorphism on $U$ .

4.1step 1.1step 1.2step 2.1step 3.1step 3.2F2F5F8algebra

By step 1.2, steps 2.1, 3.1 and 3.2 put every defining generator's coproduct in $I_{D J} \otimes T + T \otimes I_{D J}$ . This sum is a two-sided ideal: for pure tensors, left and right multiplication preserve each summand because $I_{D J}$ is a two-sided ideal; the product formula in [F5] gives these products. Since $Δ$ is an algebra map, it contains the coproduct of every element of the generated ideal $I_{D J}$ .

4.2step 1.3step 3.3F11algebra

For $K_{h}$ , both antipode products are $S (K_{h}) K_{h} = K_{- h} K_{h} = 1$ and $K_{h} S (K_{h}) = K_{h} K_{- h} = 1$ . For $E_{i}$ , they are $S (E_{i}) K_{i}^{- 1} + E_{i} = - E_{i} K_{i} K_{i}^{- 1} + E_{i} = 0$ and $E_{i} S (K_{i}^{- 1}) + S (E_{i}) = E_{i} K_{i} - E_{i} K_{i} = 0$ . For $F_{i}$ , they are $S (F_{i}) + S (K_{i}) F_{i} = - K_{i}^{- 1} F_{i} + K_{i}^{- 1} F_{i} = 0$ and $F_{i} + K_{i} S (F_{i}) = F_{i} - K_{i} K_{i}^{- 1} F_{i} = 0$ . Since $S (1) = 1$ by construction, the equations hold on the unit as well. These are the two convolution equations of [F11] on every generator and on the unit.

5.1step 1.1step 4.1step 1.3F5F7F11algebra

The maps $Δ$ and $ε$ descend by steps 4.1 and 1.3. For $K_{h}$ , both iterated coproducts equal $K_{h} \otimes K_{h} \otimes K_{h}$ . For $E_{i}$ , each equals $E_{i} \otimes K_{i}^{- 1} \otimes K_{i}^{- 1} + 1 \otimes E_{i} \otimes K_{i}^{- 1} + 1 \otimes 1 \otimes E_{i}$ ; for $F_{i}$ , each equals $F_{i} \otimes 1 \otimes 1 + K_{i} \otimes F_{i} \otimes 1 + K_{i} \otimes K_{i} \otimes F_{i}$ . Thus coassociativity holds on generators. On $K_{h}$ both counit composites equal $K_{h}$ ; on $E_{i}$ they give $0 + E_{i} = E_{i}$ and $E_{i} \cdot 1 + 0 = E_{i}$ ; on $F_{i}$ they give $0 + F_{i} = F_{i}$ and $F_{i} \cdot 1 + 0 = F_{i}$ . Each side of these identities is an algebra homomorphism, so equality on the generators and unit proves equality on all of $U$ , giving the bialgebra axioms in [F11].

5.2step 4.2F11algebra

Write $Δ (x) = \sum x_{(1)} \otimes x_{(2)}$ . If the two convolution equations hold for words $x$ and $y$ , then anti-multiplicativity of $S$ and multiplicativity of $Δ$ give $\sum S (x_{(1)} y_{(1)}) x_{(2)} y_{(2)} = \sum S (y_{(1)}) (\sum S (x_{(1)}) x_{(2)}) y_{(2)} = ε (x) ε (y) 1$ . Likewise $\sum x_{(1)} y_{(1)} S (x_{(2)} y_{(2)}) = \sum x_{(1)} (\sum y_{(1)} S (y_{(2)})) S (x_{(2)}) = ε (y) ε (x) 1$ . Since $ε$ is multiplicative and all words are products of generators, induction gives both identities on every word and hence every element of $U$ . Thus $S$ is an antipode as defined in [F11].

6.1step 3.3step 5.2F12algebra∎

Uniqueness follows from [F12]. Since $S$ reverses products, $S^{2}$ is an algebra homomorphism, and on generators $S^{2} (E_{i}) = K_{i}^{- 1} E_{i} K_{i} = q_{i}^{- 2} E_{i}$ , $S^{2} (F_{i}) = K_{i}^{- 1} F_{i} K_{i} = q_{i}^{2} F_{i}$ , and $S^{2} (K_{h}) = K_{h}$ . This proves the statement.

Source note

JKK §1, display (1.6), printed p. 6, gives exactly the coproduct, counit and antipode convention used here but asserts the Hopf structure without proving relation preservation. Berkeley Ch. 13 §13.1.3, printed pp. 308–309, gives the positive Serre quasiprimitivity/Hopf-ideal results and a different coproduct convention; the convention-sensitive relation calculations are local. The scaffold's claim that the convolution antipode equations extend from generators because all maps involved are algebra or anti-algebra homomorphisms is invalid: $m (S \otimes id) Δ$ is not generally an algebra homomorphism. Step 5.2 supplies the required word-induction argument. The free-ideal coideal claim uses the explicitly proved tensor-quotient kernel calculation in step 1.2, rather than an unproved freeness or choice argument.

Depends on

Used by

Coproduct, antipode and q-binomial expansion in U_q(sl₂) Example

Dependency tree · two levels

43 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Kyeonghoon Jeong, Seok-Jin Kang and Masaki Kashiwara, Crystal Bases for Quantum Generalized Kac-Moody Algebras, arXiv:math/0305390 (standard reference, not scraped)
Richard Borcherds, Mark Haiman, Theo Johnson-Freyd, Nicolai Reshetikhin and Vera Serganova, Berkeley Lectures on Lie Groups and Quantum Groups (standard reference, not scraped)