Matsumoto's Theorem

Matsumoto's theorem provides an explicit presentation of $K_2(F)$ for any field $F$ , showing that it is generated by Steinberg symbols subject only to the bimultiplicativity and Steinberg relations. This result is foundational for Milnor K-theory and connects algebraic K-theory to classical symbol maps in number theory.

Statement

Theorem1.2Matsumoto's Theorem

For any field $F$ , the group $K_2(F)$ admits the presentation

$K_2(F) \cong \frac{F^{\times} \otimes_{\mathbb{Z}} F^{\times}}{\langle a \otimes (1-a) : a \in F \setminus \{0, 1\} \rangle}.$

Equivalently, $K_2(F)$ is the abelian group generated by symbols $\{a, b\}$ for $a, b \in F^{\times}$ , subject to:

Bimultiplicativity: $\{ab, c\} = \{a, c\}\{b, c\}$ and $\{a, bc\} = \{a, b\}\{a, c\}$
Steinberg relation: $\{a, 1 - a\} = 1$ for all $a \in F \setminus \{0, 1\}$

The symbol map $F^{\times} \times F^{\times} \to K_2(F)$ , $(a, b) \mapsto \{a, b\}$ , is the universal Steinberg symbol.

Consequences of the symbol relations

RemarkDerived identities

From bimultiplicativity and the Steinberg relation, one derives:

Antisymmetry: $\{a, b\} = -\{b, a\}$ for all $a, b \in F^{\times}$ .

Proof: From $a + b = a(1 - (-b/a))$ , the Steinberg relation on $-b/a$ gives $\{-b/a, 1 + b/a\} = 1$ . Expanding by bimultiplicativity and simplifying yields $\{a, b\}\{b, a\} = \{-1, -1\}^0 = 1$ after careful manipulation.
$\{a, -a\} = 1$ : Set $b = -a$ in the identity $\{a, 1-a\} = 1$ applied to the element $a/(a+b)$ (when $a + b \neq 0$ ).
$\{a, a\} = \{a, -1\}$ : From $\{a, -a\} = 1$ , we get $\{a, a\}\{a, -1\} = \{a, -a\} = 1$ .
$\{-1, -1\}$ has order dividing 2: From $\{-1, -1\} = \{-1, -1\}^{-1}$ using antisymmetry.

Proof sketch

Proof

The proof that $K_2(F) = \ker(\operatorname{St}(F) \to E(F))$ is generated by Steinberg symbols proceeds in several steps.

Step 1: The symbols lie in $K_2(F)$ . For $a, b \in F^{\times}$ , define $\{a, b\} \in \operatorname{St}(F)$ using the Steinberg generators. One verifies that $\phi(\{a, b\}) = 1 \in E(F)$ , so $\{a, b\} \in K_2(F) = \ker \phi$ .

Step 2: Universality. Consider any Steinberg symbol, i.e., a bimultiplicative map $c: F^{\times} \times F^{\times} \to A$ (abelian group) satisfying $c(a, 1-a) = 0$ . We must show $c$ factors through $K_2(F)$ .

Step 3: Reduction to the Dennis--Stein symbols. Define elements $\langle a, b \rangle \in \operatorname{St}(F)$ for $a, b \in F$ with $1 + ab \neq 0$ :

$\langle a, b \rangle = x_{12}(a) \, x_{21}(b(1+ab)^{-1}) \, x_{12}(-a(1+ab)^{-1}(1+ba)) \cdot h_{12}(1+ab)^{-1}.$

These satisfy simpler relations than the Steinberg symbols and generate $K_2(F)$ .

Step 4: Presentation via Dennis--Stein symbols. Every element of $K_2(F)$ is a product of Dennis--Stein symbols. Each Dennis--Stein symbol $\langle a, b \rangle$ with $ab \neq 0$ can be expressed as a product of Steinberg symbols:

$\langle a, b \rangle = \{-(1 + ab), ab^{-1}(1+ab)^{-1}\} \cdot (\text{correction terms}).$

Step 5: Verifying the Steinberg relation. The identity $\{a, 1-a\} = 1$ in $K_2(F)$ follows from the Steinberg relations in $\operatorname{St}(F)$ : the specific element $\{a, 1-a\}$ maps to the identity matrix in $E(F)$ and can be shown to equal $1$ in $\operatorname{St}(F)$ by direct computation.

Step 6: No extra relations. This is the hardest part. One shows that any central element of $\operatorname{St}(F)$ can be expressed as a product of symbols, using the structure theory of $\operatorname{St}(F)$ as a central extension and properties of $BN$ -pairs. $\square$

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Applications

ExampleHilbert reciprocity via K₂

For a number field $F$ with places $v$ , the tame symbol maps assemble into

$K_2(F) \xrightarrow{\oplus_v \partial_v} \bigoplus_v k(v)^{\times}$

where $k(v)$ is the residue field at $v$ . Moore's reciprocity theorem states that for the composite

$K_2(F) \to \bigoplus_v K_2(F_v) \xrightarrow{\oplus \operatorname{inv}_v} \mathbb{Q}/\mathbb{Z}$

the image is zero: $\sum_v \operatorname{inv}_v(\{a, b\}_v) = 0$ for all $a, b \in F^{\times}$ . This is the K-theoretic formulation of global reciprocity.

ExampleConnection to the Brauer group

For a field $F$ containing $n$ -th roots of unity $\mu_n$ , the norm residue symbol gives a map

$K_2(F)/nK_2(F) \to \operatorname{Br}(F)[n]$

sending $\{a, b\}$ to the class of the cyclic algebra $(a, b)_n$ . The Merkurjev--Suslin theorem shows this map is an isomorphism:

$K_2(F)/nK_2(F) \cong H^2(F, \mu_n^{\otimes 2}).$

This was the first major case of the Bloch--Kato conjecture.