Definition 64.4.1. The trace of the endomorphism $a$ is the sum of the diagonal entries of a matrix representing it. This defines an additive map $\text{Tr} : \text{End}_\Lambda (\Lambda ^{\oplus m}) \to \Lambda ^\natural $.
64.4 Traces
We now explain how to take the trace of an endomorphism of a module over a noncommutative ring. Fix a finite ring $\Lambda $ with cardinality prime to $p$. Typically, $\Lambda $ is the group ring $(\mathbf{Z}/\ell ^ n\mathbf{Z})[G]$ for some finite group $G$. By convention, all the $\Lambda $-modules considered will be left $\Lambda $-modules.
We introduce the following notation: We set $\Lambda ^\natural $ to be the quotient of $\Lambda $ by its additive subgroup generated by the commutators (i.e., the elements of the form $ab-ba$, $a, b \in \Lambda $). Note that $\Lambda ^\natural $ is not a ring.
For instance, the module $(\mathbf{Z}/\ell ^ n\mathbf{Z})[G]^\natural $ is the dual of the class functions, so
\[ (\mathbf{Z}/\ell ^ n\mathbf{Z})[G]^\natural = \bigoplus \nolimits _{\text{conjugacy classes of }G} \mathbf{Z}/\ell ^ n\mathbf{Z}. \]
For a free $\Lambda $-module, we have $\text{End}_\Lambda (\Lambda ^{\oplus m}) = \text{Mat}_ n(\Lambda )$. Note that since the modules are left modules, representation of endomorphism by matrices is a right action: if $a \in \text{End}(\Lambda ^{\oplus m})$ has matrix $A$ and $v \in \Lambda $, then $a(v) = v A$.
Exercise 64.4.2. Given maps show that $\text{Tr}(ab) = \text{Tr}(ba)$.
\[ \Lambda ^{\oplus m} \xrightarrow {a} \Lambda ^{\oplus n} \quad \text{and}\quad \Lambda ^{\oplus n} \xrightarrow {b} \Lambda ^{\oplus m} \]
We extend the definition of the trace to a finite projective $\Lambda $-module $P$ and an endomorphism $\varphi $ of $P$ as follows. Write $P$ as the summand of a free $\Lambda $-module, i.e., consider maps $P \xrightarrow {a} \Lambda ^{\oplus n} \xrightarrow {b} P$ with
$\Lambda ^{\oplus n} = \mathop{\mathrm{Im}}(a) \oplus \mathop{\mathrm{Ker}}(b)$; and
$b\circ a = \text{id}_ P$.
Then we set $\text{Tr}(\varphi ) = \text{Tr}(a\varphi b)$. It is easy to check that this is well-defined, using the previous exercise.
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