Difference between revisions of "Cross product"
m (Added category TEXpartially done) |
(Link to Vector product for the special common case.) |
||
Line 1: | Line 1: | ||
{{TEX|part}} | {{TEX|part}} | ||
− | ''crossed product, of a group $G$ and a ring $K$'' | + | ''crossed product, of a [[group]] $G$ and a [[ring]] $K$'' |
+ | |||
+ | ''of a vector $a$ by a vector $b$ in $\mathbb{R}^3$, see [[Vector product]].'' | ||
An associative ring defined as follows. Suppose one is given a mapping $\sigma$ of a group $G$ into the isomorphism group of an associative ring $K$ with identity, and a family | An associative ring defined as follows. Suppose one is given a mapping $\sigma$ of a group $G$ into the isomorphism group of an associative ring $K$ with identity, and a family |
Revision as of 07:24, 2 June 2016
crossed product, of a group $G$ and a ring $K$
of a vector $a$ by a vector $b$ in $\mathbb{R}^3$, see Vector product.
An associative ring defined as follows. Suppose one is given a mapping $\sigma$ of a group $G$ into the isomorphism group of an associative ring $K$ with identity, and a family
$$ \rho = \{ \rho_{g,h} | g,h \in G\} $$
of invertible elements of $K$, satisfying the conditions
$$ \rho_{g_1,g_2}\rho_{g_1g_2,g_3} = \rho^{\sigma(g_1)}_{g_2,g_3}\rho_{g_1,g_2g_3} $$ $$ \alpha^{\sigma(g_2)\sigma(g_1)} = \rho_{g_1,g_2}\alpha^{\sigma(g_1g_2)}\rho^{-1}_{g_1,g_2} $$
for all $\alpha\in K$ and $g_1,g_2,g_3\in G$. The family $\rho$ is called a factor system. Then the cross product of $G$ and $K$ with respect to the factor system $\rho$ and the mapping $\sigma$ is the set of all formal finite sums of the form
$$ \sum_{g\in G} \alpha_g t_g $$
where $\alpha_g \in K$ and the $t_g$ are symbols uniquely assigned to every element $g\in G$, with binary operations defined by
$$ \sum_{g\in G} \alpha_g t_g + \sum_{g\in G} \beta_g t_g = \sum_{g\in G} (\alpha_g+\beta_g)t_g,$$ $$ \left(\sum_{g\in G}\alpha_gt_g\right) \left(\sum_{g\in G}\beta_gt_g\right) = \sum_{g\in G} \left(\sum_{h_1h_2=g}\alpha_{h_1}\beta^{\sigma(h_1)}_{h_2}\rho_{h_1,h_2}\right) t_g $$
This ring is denoted by $K(G, \rho, \sigma)$; the elements $t_g$ form a $K$-basis for the ring.
If $\sigma$ maps $G$ onto the identity automorphism of $K$, then $K(G, \rho)$ is called a twisted or crossed group ring, and if, in addition, $\rho_{g,h}=1$ for all $g,h\in G$, then $K(G,\rho,\sigma)$ is the group ring of $G$ over $K$ (see Group algebra).
Let $K$ be a field and $\sigma$ a monomorphism. Then $K(G,\rho,\sigma)$ is a simple ring, being the cross product of the field with its Galois group.
References
[1] | S.K. Sehgal, "Topics in group rings" , M. Dekker (1978) |
[2] | A.A. Bovdi, "Cross products of semi-groups and rings" Sibirsk. Mat. Zh. , 4 (1963) pp. 481–499 (In Russian) |
[3] | A.E. Zalesskii, A.V. Mikhalev, "Group rings" J. Soviet Math. , 4 (1975) pp. 1–74 Itogi Nauk. i Tekhn. Sovrem. Probl. Mat. , 2 (1973) pp. 5–118 |
[4] | D.S. Passman, "The algebraic structure of group rings" , Wiley (1977) |
Comments
In the defining relations for a factor system above , e.g., of course stands for the result of applying the automorphism to the element . If for all , then one obtains the skew group ring . Cross products arise naturally when dealing with extensions. Indeed, let be a normal subgroup of . Choose a set of representatives of in . Then every , the group algebra of , can be written as a unique sum , . Now write
Then the define a factor system (for the group and the ring relative to the set of automorphisms ) and
Up to Brauer equivalence every central simple algebra is a cross product, but not every division algebra is isomorphic to a cross product. Two algebras over are Brauer equivalent if is isomorphic to for suitable and . Here is the algebra of matrices over . Cf. also Brauer group.
Cross product. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Cross_product&oldid=38904