We can ask if the group is determined by the group ring. The answer is“no” if the group ring is considered as an associative algebra: for example,the complex group ring of a finite abelian group is just a sum of copies of Ccorresponding to the irreducible representations, so two finite abelian groupshave isomorphic group rings if and only if they have the same order. Howeverthe group can be recovered from the Hopf algebra as the group-like elements:Definition 38 An element of a Hopf algebra is called group-like if ∆(a) = a⊗aand (a) = 1.Exercise 39 Show that the group-like elements of a Hopf algebra form a group.Show that the group-like elements of a group ring of G form a group that can b enaturally identified with the group G. What is the group of group-like elementsof the universal enveloping algebra of a Lie algebra?The universal enveloping algebra is not the only analogue of the group ringfor a Lie group. Another analogue, often used in analysis, is the algebra ofcontinuous (or smooth) functions of compact support (or L1functions, or finitemeasures...) under convolution. For p-adic groups one can take the lo callyconstant functions with compact support. These algebras look more like thegroup algebra of a finite group, but is less convenient in some ways as they neednot have a coproduct.Example 40 The main point of all this is that the UEA of a Lie algebra is aHopf algebra. In other words it behaves as if it were a group or group ring, whichis of course an approximation to the Lie group of the Lie algebra. The map fromUg to Ug⊗Ug is given by the Leibniz formula g 7→ 1⊗g+g×1 from calculus. Thisis really just the formula telling you how to differentiate a product:ddx(fg) =ddx(f)g + fddx(g), where you can see the mapddx7→ddx⊗ 1 + 1 ⊗ddx.The fact that the map ∆ extends to a ring homomorphism follows from theuniversal property of the UEA.We would like to reconstruct the Lie algebra from its universal envelopingalgebra in the same way we reconstructed a group from its group algebra.Definition 41 An element of a Hopf algebra is called primitive if it satisfiesthe Leibniz identity ∆(a) = a ⊗ 1 + 1 ⊗ a.Exercise 42 Show that the primitive elements of a Hopf algebra form a Liealgebra.A natural guess is that the Lie algebra consists of the primitive elements of theuniversal enveloping algebra. This fails over fields of positive characteristic p:Exercise 43 If a is primitive in a Hopf algebra of prime characteristic p > 0,so is ap. Find the Lie algebra of primitive elements of the universal envelopingalgebra of a 1-dimensional Lie algebra over the finite field Fp.Sometimes in characteristic p one works with restricted Lie algebras: these areLie algebras together with a “p’th power operation” a 7→ a[p] behaving like thep-th power of derivations.25Lemma 44 Over a field of characteristic 0, an element of the UEA is primitiveif and only if it is in the Lie algebra.Proof It is obvious that elements of the Lie algebra are primitive, so we needto show that primitive elements are in the Lie algebra. By the PBW theorem,the coalgebra structures on any two Lie algebras of the same dimension areisomorphic, so we can just the lemma for one Lie algebra of any dimension,say the abelian one. But in this case the dual of the coalgebra is a ring ofpower series (since we are in char 0). Primitive elements have to vanish on alldecomposable elements, which are all elements of degree at least 2, so primitiveelements have degree 1 and are therefore in the Lie algebra. The proof failsin positive characteristic because the dual algebra of the coalgebra of a UEA isno longer a power series ring. If the coalgebra is Z[D] then the dual algebrais Z[{xi/i!}], so reducing mod p we get an algebra generated by elements ofdegrees pneach of which has p’th power 0. This is another indication that theUEA is the wrong object if we are not working over fields of char 0.There is a second way to associate a Hopf algebra to a Lie group, which is insome sense dual to the UEA, which is to take the ring of polynomial functionson an (algebraic) Lie group. The UEA consists of (left invariant) differentialoperators and is cocommutative but not usually commutative, while the ringof (polynomial) functions is commutative but not usually cocommutative. Itworks as follows: suppose that G is an algebraic group contained in Rn. Thenit has a coordinate ring O(G) = R[x1. · · · , xn]/(I) where the ideal I is thepolynomials vanishing on R. The product map G × G 7→ G induces a dual mapO(G) 7→ O(G) ⊗ O(G), and similarly the unit of G induces a map O(G) 7→ R,so we have all the data for a commutative Hopf algebra.Example 45 Suppose G is the general linear group SL2(R). Then the coordi-nate ring is O(G) = R[a, b, c, d]/(ad − bc − 1). The coproduct is given by thegroup product:a1b1c1d1a2b2c2d2=a1a2+ b1c2a1b2+ b1d2c1a2+ d1c2c1b2+ d1d2so ∆(a) = a⊗a+b⊗c, ∆(b) = a⊗b+b⊗d, ∆(c) = c⊗a +d⊗b, ∆(d) = c⊗b+d⊗d.The counit is given by η(a) = 1, η(b) = 0, η(c) = 0, η(d) = 1. The antipode isS(a) = d, S(b) = −b, S(c) = −c, S(d) = a.Example 46 Following Quillen and Milnor, we will show that the Steenrodalgebra in algebraic topology is a sort of infinite dimensional Lie group.First recall the original definition of the Steenrod algebra. The Steenrodalgebra is the algebra of stable cohomology operations mod 2, so can be found bycalculating the cohomology of Eilenberg–Maclane spaces, which was originallydone by H. Cartan and Serre. They showed that the Steenrod algebra is thealgebra over F2generated by elements Sqqfor q = 1, 2, 3 · · · modulo the AdemrelationsSqiSqj=[i/2]Xk=0j − k − 1i − 2kSqi+j−kSqk26which not even algebraic topologists are able to remember. Cartan also gave aformula for the action of Steenrod squares on a cup productSqn(x ∪ y) =Xi+j=n(Sqix) ∪ (Sqjy)which we know should be interpreted as a coproduct on the Steenrod algebra∆Sqn=Xi+j=nSqi⊗ SqjIn other words the Steenrod algebra is a cocommutative Hopf algebra over thefield with 2 elements. A cocommutative Hopf algebra should be thought of assomething like a (universal enveloping algebra of a Lie) group, so should berelated to the automorphisms of something. We will show that the Steenrodalgebra is in some sense the automorphism group of the 1-dimensional additivegroup.At first sight this makes no sense. The automorphism group of the additiveLie group R is just R∗which looks