Editing Trace (linear algebra) (section)

== Lie algebra ==
The trace is a map of Lie algebras <math>\operatorname{tr}:\mathfrak{gl}_n\to K</math> from the Lie algebra <math>\mathfrak{gl}_n</math> of linear operators on an {{mvar|n}}-dimensional space ({{math|''n'' × ''n''}} matrices with entries in <math>K</math>) to the Lie algebra {{mvar|K}} of scalars; as {{mvar|K}} is Abelian (the Lie bracket vanishes), the fact that this is a map of Lie algebras is exactly the statement that the trace of a bracket vanishes:
<math display="block">\operatorname{tr}([\mathbf{A}, \mathbf{B}]) = 0 \text{ for each }\mathbf A,\mathbf B\in\mathfrak{gl}_n.</math>

The kernel of this map, a matrix whose trace is [[0 (number)|zero]], is often said to be '''{{visible anchor|traceless}}''' or '''{{visible anchor|trace free}}''', and these matrices form the [[simple Lie algebra]] <math>\mathfrak{sl}_n</math>, which is the [[Lie algebra]] of the [[special linear group]] of matrices with determinant 1. The special linear group consists of the matrices which do not change volume, while the [[special linear Lie algebra]] is the matrices which do not alter volume of ''infinitesimal'' sets.

In fact, there is an internal [[direct sum of Lie algebras|direct sum]] decomposition <math>\mathfrak{gl}_n = \mathfrak{sl}_n \oplus K</math> of operators/matrices into traceless operators/matrices and scalars operators/matrices. The projection map onto scalar operators can be expressed in terms of the trace, concretely as:
<math display="block">\mathbf{A} \mapsto \frac{1}{n}\operatorname{tr}(\mathbf{A})\mathbf{I}.</math>

Formally, one can compose the trace (the [[counit]] map) with the unit map <math>K\to\mathfrak{gl}_n</math> of "inclusion of [[scalar transformation|scalars]]" to obtain a map <math>\mathfrak{gl}_n\to\mathfrak{gl}_n</math> mapping onto scalars, and multiplying by {{mvar|n}}. Dividing by {{mvar|n}} makes this a projection, yielding the formula above.

In terms of [[short exact sequence]]s, one has
<math display="block">0 \to \mathfrak{sl}_n \to \mathfrak{gl}_n \overset{\operatorname{tr}}{\to} K \to 0</math>
which is analogous to
<math display="block">1 \to \operatorname{SL}_n \to \operatorname{GL}_n \overset{\det}{\to} K^* \to 1</math>
(where <math>K^*=K\setminus\{0\}</math>) for [[Lie group]]s. However, the trace splits naturally (via <math>1/n</math> times scalars) so <math>\mathfrak{gl}_n=\mathfrak{sl}_n\oplus K</math>, but the splitting of the determinant would be as the {{mvar|n}}th root times scalars, and this does not in general define a function, so the determinant does not split and the general linear group does not decompose:
<math display="block">\operatorname{GL}_n \neq \operatorname{SL}_n \times K^*.</math>

=== Bilinear forms ===
The [[bilinear form]] (where {{math|'''X'''}}, {{math|'''Y'''}} are square matrices)
<math display="block">B(\mathbf{X}, \mathbf{Y}) = \operatorname{tr}(\operatorname{ad}(\mathbf{X})\operatorname{ad}(\mathbf{Y}))</math>
: where <math>\operatorname{ad}(\mathbf{X})\mathbf{Y} = [\mathbf{X}, \mathbf{Y}] = \mathbf{X}\mathbf{Y} - \mathbf{Y}\mathbf{X}</math>
: and for orientation, if <math>\operatorname{det} \mathbf{Y} \ne 0 </math>
:: then <math>\operatorname{ad}(\mathbf{X}) = \mathbf{X} - \mathbf{Y}\mathbf{X}\mathbf{Y}^{-1} ~.</math>

<math> B(\mathbf{X}, \mathbf{Y})</math> is called the [[Killing form]]; it is used to classify [[Lie algebra]]s.

The trace defines a bilinear form:
<math display="block">(\mathbf{X}, \mathbf{Y}) \mapsto \operatorname{tr}(\mathbf{X}\mathbf{Y}) ~.</math>

The form is symmetric, non-degenerate<ref group=note>This follows from the fact that {{math|1=tr('''A'''*'''A''') = 0}} [[if and only if]] {{math|1='''A''' = '''0'''}}.</ref> and associative in the sense that:
<math display="block">\operatorname{tr}(\mathbf{X}[\mathbf{Y}, \mathbf{Z}]) = \operatorname{tr}([\mathbf{X}, \mathbf{Y}]\mathbf{Z}).</math>

For a complex simple Lie algebra (such as {{math|<math>\mathfrak{sl}</math><sub>''n''</sub>}}), every such bilinear form is proportional to each other; in particular, to the Killing form{{Citation needed|reason=Either a source or proof is needed|date=June 2022}}.

Two matrices {{math|'''X'''}} and {{math|'''Y'''}} are said to be ''trace orthogonal'' if
<math display="block">\operatorname{tr}(\mathbf{X}\mathbf{Y}) = 0.</math>

There is a generalization to a general representation <math>(\rho,\mathfrak{g},V)</math> of a Lie algebra <math>\mathfrak{g}</math>, such that <math>\rho</math> is a homomorphism of Lie algebras <math>\rho: \mathfrak{g} \rightarrow \text{End}(V).</math> The trace form <math>\text{tr}_V</math> on <math>\text{End}(V)</math> is defined as above. The bilinear form
<math display="block">\phi(\mathbf{X},\mathbf{Y}) = \text{tr}_V(\rho(\mathbf{X})\rho(\mathbf{Y}))</math>
is symmetric and invariant due to cyclicity.