What relations do elementary matrices secretly satisfy?
In Guide 2 we used the Steinberg relations — identities the elementary matrices e_ij(r) obviously satisfy. Now invert the question. Write down the abstract group St(R) generated by symbols x_ij(r) subject to *only* those obvious relations, then map it onto E(R) by x_ij(r) ↦ e_ij(r). If the obvious relations were all the relations, this map would be an isomorphism. K_2 is precisely the failure: K_2(R) = ker(St(R) → E(R)). It is the group of hidden relations among row operations.
Steinberg group St(R): generators x_ij(r), i != j, r in R, relations
(S1) x_ij(r) x_ij(s) = x_ij(r+s)
(S2) [ x_ij(r), x_kl(s) ] = 1 if j != k and i != l
(S3) [ x_ij(r), x_jk(s) ] = x_ik(rs) if i != k
There is a surjection phi: St(R) -> E(R), x_ij(r) |-> e_ij(r).
Definition: K_2(R) := ker( phi ) = ker( St(R) -> E(R) ).
Key structural fact (Milnor):
1 -> K_2(R) -> St(R) -> E(R) -> 1
is a CENTRAL extension, and in fact the UNIVERSAL central extension
of the perfect group E(R). Hence
K_2(R) = center of St(R) = H_2(E(R); Z),
the Schur multiplier of E(R).Steinberg symbols you can compute
K_2 is abstract until you have an element in your hands. For a field k, every class in K_2(k) is built from Steinberg symbols {u, v}, defined for units u, v ∈ k^×. They are bilinear and satisfy one striking extra relation that gives the theory its flavour.
Steinberg symbol {u, v} in K_2(k), for u, v in k^x. Rules:
(bi-1) {u u', v} = {u, v}{u', v}
(bi-2) {u, v v'} = {u, v}{u, v'}
(STEINBERG) {u, 1 - u} = 1 whenever u != 0, 1.
Consequences you can derive in two lines:
* {u, -u} = 1. (since -u = (1-u)/(1-u^{-1}), then expand)
* {u, v} = {v, u}^{-1} (skew-symmetry)
* {u, u} = {u, -1} (a 2-torsion-flavored identity)
Matsumoto's theorem: for a field k,
K_2(k) = (k^x (x) k^x) / < u (x) (1-u) >,
the free thing on symbols modulo exactly the Steinberg relation.
Worked value: K_2(F_q) = 0 for every finite field F_q
(every symbol is trivial because k^x is cyclic and the relation bites).The relation {u, 1 − u} = 1 should look uncanny: it is the K-theoretic incarnation of the Steinberg relation and, astonishingly, the same algebra controls the dilogarithm, tame symbols in number theory, and Hilbert symbols. Generalizing only these rules to all degrees gives Milnor K-theory K^M_n(k) = (k^×)^⊗n modulo Steinberg relations — a hands-on theory that, by the Bloch–Kato theorem, computes Galois cohomology.
Quillen's leap to all higher K
We now have K_0, K_1, K_2 by three ad hoc constructions. The miracle, due to Quillen, is that all of them — and infinitely many higher K-groups K_n — come from a single space. Take the classifying space BGL(R), which is connected but has the wrong, nonabelian fundamental group. The plus construction surgically kills the perfect subgroup E(R) ⊆ π_1 without changing homology, producing a space BGL(R)^+. Define K_n(R) = π_n(BGL(R)^+) for n ≥ 1.
- Check the new definition reproduces the old ones: π_1(BGL^+) = GL/E = K_1, and π_2(BGL^+) = H_2(E) = K_2. Consistency holds.
- Now K_n for all n ≥ 1 is simply a higher homotopy group — abelian for free, since π_n of a space is abelian for n ≥ 2.
- The localization sequences of Guide 3 are now genuine long exact sequences of homotopy groups, extending upward forever.
Quillen immediately computed the K-theory of finite fields: K_{2i-1}(F_q) ≅ Z/(q^i − 1) and K_{2i}(F_q) = 0 for i ≥ 1. That a purely algebraic ring has K-groups indexed like the homotopy groups of a space — and computable — is the whole point. The higher K-groups of Z are still not all known; computing them is bound up with deep number theory, which is where Guide 5 heads.