Кіріспе
Супералгебралар теориясында, егер A коммутативті супералгебра болса, V – еркін оң жақтағы A супермодулі болса және T – V-ден өзіне ендоморфизм болса, онда T супертрегі, str(T) келесі із диаграммасымен анықталады:
In the theory of superalgebras, if A is a commutative superalgebra, V is a free right A supermodule and T is an endomorphism from V to itself, then the supertrace of T, str(T) is defined by the following trace diagram:
More concretely, if we write out T in block matrix form after the decomposition into even and odd subspaces as follows,
then the supertrace
str(T) = the ordinary trace of T00 − the ordinary trace of T11. Let us show that the supertrace does not depend on a basis. Suppose e1, , ep are the even basis vectors and ep+1, , ep+q are the odd basis vectors. Then, the components of T, which are elements of A, are defined as
The grading of Tij is the sum of the gradings of T, ei, ej mod 2. A change of basis to e1', , ep', e(p+1)', , e(p+q)' is given by the supermatrix
and the inverse supermatrix
where of course, AA−1 = A−1A = 1 (the identity). We can now check explicitly that the supertrace is basis independent. In the case where T is even, we have
In the case where T is odd, we have
The ordinary trace is not basis independent, so the appropriate trace to use in the Z2 graded setting is the supertrace. The supertrace satisfies the property
for all T1, T2 in End(V). In particular, the supertrace of a supercommutator is zero. In fact, one can define a supertrace more generally for any associative superalgebra E over a commutative superalgebra A as a linear map tr: E > A which vanishes on supercommutators. Such a supertrace is not uniquely defined; it can always at least be modified by multiplication by an element of A.
More concretely, if we write out T in block matrix form after the decomposition into even and odd subspaces as follows,
then the supertrace
str(T) = the ordinary trace of T00 − the ordinary trace of T11. Let us show that the supertrace does not depend on a basis. Suppose e1, , ep are the even basis vectors and ep+1, , ep+q are the odd basis vectors. Then, the components of T, which are elements of A, are defined as
The grading of Tij is the sum of the gradings of T, ei, ej mod 2. A change of basis to e1', , ep', e(p+1)', , e(p+q)' is given by the supermatrix
and the inverse supermatrix
where of course, AA−1 = A−1A = 1 (the identity). We can now check explicitly that the supertrace is basis independent. In the case where T is even, we have
In the case where T is odd, we have
The ordinary trace is not basis independent, so the appropriate trace to use in the Z2 graded setting is the supertrace. The supertrace satisfies the property
for all T1, T2 in End(V). In particular, the supertrace of a supercommutator is zero. In fact, one can define a supertrace more generally for any associative superalgebra E over a commutative superalgebra A as a linear map tr: E > A which vanishes on supercommutators. Such a supertrace is not uniquely defined; it can always at least be modified by multiplication by an element of A.
Нақтырақ айтқанда, егер біз T-ді жұп және тақ субкеңістіктерге жіктеуден кейін блок матрица түрінде жазсақ, онда супертрек:
In the theory of superalgebras, if A is a commutative superalgebra, V is a free right A supermodule and T is an endomorphism from V to itself, then the supertrace of T, str(T) is defined by the following trace diagram:
More concretely, if we write out T in block matrix form after the decomposition into even and odd subspaces as follows,
then the supertrace
str(T) = the ordinary trace of T00 − the ordinary trace of T11. Let us show that the supertrace does not depend on a basis. Suppose e1, , ep are the even basis vectors and ep+1, , ep+q are the odd basis vectors. Then, the components of T, which are elements of A, are defined as
The grading of Tij is the sum of the gradings of T, ei, ej mod 2. A change of basis to e1', , ep', e(p+1)', , e(p+q)' is given by the supermatrix
and the inverse supermatrix
where of course, AA−1 = A−1A = 1 (the identity). We can now check explicitly that the supertrace is basis independent. In the case where T is even, we have
In the case where T is odd, we have
The ordinary trace is not basis independent, so the appropriate trace to use in the Z2 graded setting is the supertrace. The supertrace satisfies the property
for all T1, T2 in End(V). In particular, the supertrace of a supercommutator is zero. In fact, one can define a supertrace more generally for any associative superalgebra E over a commutative superalgebra A as a linear map tr: E > A which vanishes on supercommutators. Such a supertrace is not uniquely defined; it can always at least be modified by multiplication by an element of A.
More concretely, if we write out T in block matrix form after the decomposition into even and odd subspaces as follows,
then the supertrace
str(T) = the ordinary trace of T00 − the ordinary trace of T11. Let us show that the supertrace does not depend on a basis. Suppose e1, , ep are the even basis vectors and ep+1, , ep+q are the odd basis vectors. Then, the components of T, which are elements of A, are defined as
The grading of Tij is the sum of the gradings of T, ei, ej mod 2. A change of basis to e1', , ep', e(p+1)', , e(p+q)' is given by the supermatrix
and the inverse supermatrix
where of course, AA−1 = A−1A = 1 (the identity). We can now check explicitly that the supertrace is basis independent. In the case where T is even, we have
In the case where T is odd, we have
The ordinary trace is not basis independent, so the appropriate trace to use in the Z2 graded setting is the supertrace. The supertrace satisfies the property
for all T1, T2 in End(V). In particular, the supertrace of a supercommutator is zero. In fact, one can define a supertrace more generally for any associative superalgebra E over a commutative superalgebra A as a linear map tr: E > A which vanishes on supercommutators. Such a supertrace is not uniquely defined; it can always at least be modified by multiplication by an element of A.
str(T) = T00 ізi – T11 ізi. Супертрек негізге тәуелді емес екенін көрсетейік. e1, …, ep – жұп негіз векторлары, ал ep+1, …, ep+q – тақ негіз векторлары деп есептейік. Содан кейін T компоненттері, T элементтері болып табылатын A элементтері, былай анықталады:
In the theory of superalgebras, if A is a commutative superalgebra, V is a free right A supermodule and T is an endomorphism from V to itself, then the supertrace of T, str(T) is defined by the following trace diagram:
More concretely, if we write out T in block matrix form after the decomposition into even and odd subspaces as follows,
then the supertrace
str(T) = the ordinary trace of T00 − the ordinary trace of T11. Let us show that the supertrace does not depend on a basis. Suppose e1, , ep are the even basis vectors and ep+1, , ep+q are the odd basis vectors. Then, the components of T, which are elements of A, are defined as
The grading of Tij is the sum of the gradings of T, ei, ej mod 2. A change of basis to e1', , ep', e(p+1)', , e(p+q)' is given by the supermatrix
and the inverse supermatrix
where of course, AA−1 = A−1A = 1 (the identity). We can now check explicitly that the supertrace is basis independent. In the case where T is even, we have
In the case where T is odd, we have
The ordinary trace is not basis independent, so the appropriate trace to use in the Z2 graded setting is the supertrace. The supertrace satisfies the property
for all T1, T2 in End(V). In particular, the supertrace of a supercommutator is zero. In fact, one can define a supertrace more generally for any associative superalgebra E over a commutative superalgebra A as a linear map tr: E > A which vanishes on supercommutators. Such a supertrace is not uniquely defined; it can always at least be modified by multiplication by an element of A.
More concretely, if we write out T in block matrix form after the decomposition into even and odd subspaces as follows,
then the supertrace
str(T) = the ordinary trace of T00 − the ordinary trace of T11. Let us show that the supertrace does not depend on a basis. Suppose e1, , ep are the even basis vectors and ep+1, , ep+q are the odd basis vectors. Then, the components of T, which are elements of A, are defined as
The grading of Tij is the sum of the gradings of T, ei, ej mod 2. A change of basis to e1', , ep', e(p+1)', , e(p+q)' is given by the supermatrix
and the inverse supermatrix
where of course, AA−1 = A−1A = 1 (the identity). We can now check explicitly that the supertrace is basis independent. In the case where T is even, we have
In the case where T is odd, we have
The ordinary trace is not basis independent, so the appropriate trace to use in the Z2 graded setting is the supertrace. The supertrace satisfies the property
for all T1, T2 in End(V). In particular, the supertrace of a supercommutator is zero. In fact, one can define a supertrace more generally for any associative superalgebra E over a commutative superalgebra A as a linear map tr: E > A which vanishes on supercommutators. Such a supertrace is not uniquely defined; it can always at least be modified by multiplication by an element of A.
Tij жіктелуі T, ei, ej mod 2 жіктелімдерінің қосындысы болып табылады. Негізді e1', …, ep', e(p+1)', …, e(p+q)'-ға өзгерту суперматрицамен беріледі:
In the theory of superalgebras, if A is a commutative superalgebra, V is a free right A supermodule and T is an endomorphism from V to itself, then the supertrace of T, str(T) is defined by the following trace diagram:
More concretely, if we write out T in block matrix form after the decomposition into even and odd subspaces as follows,
then the supertrace
str(T) = the ordinary trace of T00 − the ordinary trace of T11. Let us show that the supertrace does not depend on a basis. Suppose e1, , ep are the even basis vectors and ep+1, , ep+q are the odd basis vectors. Then, the components of T, which are elements of A, are defined as
The grading of Tij is the sum of the gradings of T, ei, ej mod 2. A change of basis to e1', , ep', e(p+1)', , e(p+q)' is given by the supermatrix
and the inverse supermatrix
where of course, AA−1 = A−1A = 1 (the identity). We can now check explicitly that the supertrace is basis independent. In the case where T is even, we have
In the case where T is odd, we have
The ordinary trace is not basis independent, so the appropriate trace to use in the Z2 graded setting is the supertrace. The supertrace satisfies the property
for all T1, T2 in End(V). In particular, the supertrace of a supercommutator is zero. In fact, one can define a supertrace more generally for any associative superalgebra E over a commutative superalgebra A as a linear map tr: E > A which vanishes on supercommutators. Such a supertrace is not uniquely defined; it can always at least be modified by multiplication by an element of A.
More concretely, if we write out T in block matrix form after the decomposition into even and odd subspaces as follows,
then the supertrace
str(T) = the ordinary trace of T00 − the ordinary trace of T11. Let us show that the supertrace does not depend on a basis. Suppose e1, , ep are the even basis vectors and ep+1, , ep+q are the odd basis vectors. Then, the components of T, which are elements of A, are defined as
The grading of Tij is the sum of the gradings of T, ei, ej mod 2. A change of basis to e1', , ep', e(p+1)', , e(p+q)' is given by the supermatrix
and the inverse supermatrix
where of course, AA−1 = A−1A = 1 (the identity). We can now check explicitly that the supertrace is basis independent. In the case where T is even, we have
In the case where T is odd, we have
The ordinary trace is not basis independent, so the appropriate trace to use in the Z2 graded setting is the supertrace. The supertrace satisfies the property
for all T1, T2 in End(V). In particular, the supertrace of a supercommutator is zero. In fact, one can define a supertrace more generally for any associative superalgebra E over a commutative superalgebra A as a linear map tr: E > A which vanishes on supercommutators. Such a supertrace is not uniquely defined; it can always at least be modified by multiplication by an element of A.
және кері суперматрицамен:
In the theory of superalgebras, if A is a commutative superalgebra, V is a free right A supermodule and T is an endomorphism from V to itself, then the supertrace of T, str(T) is defined by the following trace diagram:
More concretely, if we write out T in block matrix form after the decomposition into even and odd subspaces as follows,
then the supertrace
str(T) = the ordinary trace of T00 − the ordinary trace of T11. Let us show that the supertrace does not depend on a basis. Suppose e1, , ep are the even basis vectors and ep+1, , ep+q are the odd basis vectors. Then, the components of T, which are elements of A, are defined as
The grading of Tij is the sum of the gradings of T, ei, ej mod 2. A change of basis to e1', , ep', e(p+1)', , e(p+q)' is given by the supermatrix
and the inverse supermatrix
where of course, AA−1 = A−1A = 1 (the identity). We can now check explicitly that the supertrace is basis independent. In the case where T is even, we have
In the case where T is odd, we have
The ordinary trace is not basis independent, so the appropriate trace to use in the Z2 graded setting is the supertrace. The supertrace satisfies the property
for all T1, T2 in End(V). In particular, the supertrace of a supercommutator is zero. In fact, one can define a supertrace more generally for any associative superalgebra E over a commutative superalgebra A as a linear map tr: E > A which vanishes on supercommutators. Such a supertrace is not uniquely defined; it can always at least be modified by multiplication by an element of A.
More concretely, if we write out T in block matrix form after the decomposition into even and odd subspaces as follows,
then the supertrace
str(T) = the ordinary trace of T00 − the ordinary trace of T11. Let us show that the supertrace does not depend on a basis. Suppose e1, , ep are the even basis vectors and ep+1, , ep+q are the odd basis vectors. Then, the components of T, which are elements of A, are defined as
The grading of Tij is the sum of the gradings of T, ei, ej mod 2. A change of basis to e1', , ep', e(p+1)', , e(p+q)' is given by the supermatrix
and the inverse supermatrix
where of course, AA−1 = A−1A = 1 (the identity). We can now check explicitly that the supertrace is basis independent. In the case where T is even, we have
In the case where T is odd, we have
The ordinary trace is not basis independent, so the appropriate trace to use in the Z2 graded setting is the supertrace. The supertrace satisfies the property
for all T1, T2 in End(V). In particular, the supertrace of a supercommutator is zero. In fact, one can define a supertrace more generally for any associative superalgebra E over a commutative superalgebra A as a linear map tr: E > A which vanishes on supercommutators. Such a supertrace is not uniquely defined; it can always at least be modified by multiplication by an element of A.
мұнда, әрине, AA−1 = A−1A = 1 (бірлік матрица). Енді біз супертректің негізге тәуелсіз екенін нақты тексеруге болады. Егер T жұп болса, онда:
In the theory of superalgebras, if A is a commutative superalgebra, V is a free right A supermodule and T is an endomorphism from V to itself, then the supertrace of T, str(T) is defined by the following trace diagram:
More concretely, if we write out T in block matrix form after the decomposition into even and odd subspaces as follows,
then the supertrace
str(T) = the ordinary trace of T00 − the ordinary trace of T11. Let us show that the supertrace does not depend on a basis. Suppose e1, , ep are the even basis vectors and ep+1, , ep+q are the odd basis vectors. Then, the components of T, which are elements of A, are defined as
The grading of Tij is the sum of the gradings of T, ei, ej mod 2. A change of basis to e1', , ep', e(p+1)', , e(p+q)' is given by the supermatrix
and the inverse supermatrix
where of course, AA−1 = A−1A = 1 (the identity). We can now check explicitly that the supertrace is basis independent. In the case where T is even, we have
In the case where T is odd, we have
The ordinary trace is not basis independent, so the appropriate trace to use in the Z2 graded setting is the supertrace. The supertrace satisfies the property
for all T1, T2 in End(V). In particular, the supertrace of a supercommutator is zero. In fact, one can define a supertrace more generally for any associative superalgebra E over a commutative superalgebra A as a linear map tr: E > A which vanishes on supercommutators. Such a supertrace is not uniquely defined; it can always at least be modified by multiplication by an element of A.
More concretely, if we write out T in block matrix form after the decomposition into even and odd subspaces as follows,
then the supertrace
str(T) = the ordinary trace of T00 − the ordinary trace of T11. Let us show that the supertrace does not depend on a basis. Suppose e1, , ep are the even basis vectors and ep+1, , ep+q are the odd basis vectors. Then, the components of T, which are elements of A, are defined as
The grading of Tij is the sum of the gradings of T, ei, ej mod 2. A change of basis to e1', , ep', e(p+1)', , e(p+q)' is given by the supermatrix
and the inverse supermatrix
where of course, AA−1 = A−1A = 1 (the identity). We can now check explicitly that the supertrace is basis independent. In the case where T is even, we have
In the case where T is odd, we have
The ordinary trace is not basis independent, so the appropriate trace to use in the Z2 graded setting is the supertrace. The supertrace satisfies the property
for all T1, T2 in End(V). In particular, the supertrace of a supercommutator is zero. In fact, one can define a supertrace more generally for any associative superalgebra E over a commutative superalgebra A as a linear map tr: E > A which vanishes on supercommutators. Such a supertrace is not uniquely defined; it can always at least be modified by multiplication by an element of A.
Егер T тақ болса, онда:
In the theory of superalgebras, if A is a commutative superalgebra, V is a free right A supermodule and T is an endomorphism from V to itself, then the supertrace of T, str(T) is defined by the following trace diagram:
More concretely, if we write out T in block matrix form after the decomposition into even and odd subspaces as follows,
then the supertrace
str(T) = the ordinary trace of T00 − the ordinary trace of T11. Let us show that the supertrace does not depend on a basis. Suppose e1, , ep are the even basis vectors and ep+1, , ep+q are the odd basis vectors. Then, the components of T, which are elements of A, are defined as
The grading of Tij is the sum of the gradings of T, ei, ej mod 2. A change of basis to e1', , ep', e(p+1)', , e(p+q)' is given by the supermatrix
and the inverse supermatrix
where of course, AA−1 = A−1A = 1 (the identity). We can now check explicitly that the supertrace is basis independent. In the case where T is even, we have
In the case where T is odd, we have
The ordinary trace is not basis independent, so the appropriate trace to use in the Z2 graded setting is the supertrace. The supertrace satisfies the property
for all T1, T2 in End(V). In particular, the supertrace of a supercommutator is zero. In fact, one can define a supertrace more generally for any associative superalgebra E over a commutative superalgebra A as a linear map tr: E > A which vanishes on supercommutators. Such a supertrace is not uniquely defined; it can always at least be modified by multiplication by an element of A.
More concretely, if we write out T in block matrix form after the decomposition into even and odd subspaces as follows,
then the supertrace
str(T) = the ordinary trace of T00 − the ordinary trace of T11. Let us show that the supertrace does not depend on a basis. Suppose e1, , ep are the even basis vectors and ep+1, , ep+q are the odd basis vectors. Then, the components of T, which are elements of A, are defined as
The grading of Tij is the sum of the gradings of T, ei, ej mod 2. A change of basis to e1', , ep', e(p+1)', , e(p+q)' is given by the supermatrix
and the inverse supermatrix
where of course, AA−1 = A−1A = 1 (the identity). We can now check explicitly that the supertrace is basis independent. In the case where T is even, we have
In the case where T is odd, we have
The ordinary trace is not basis independent, so the appropriate trace to use in the Z2 graded setting is the supertrace. The supertrace satisfies the property
for all T1, T2 in End(V). In particular, the supertrace of a supercommutator is zero. In fact, one can define a supertrace more generally for any associative superalgebra E over a commutative superalgebra A as a linear map tr: E > A which vanishes on supercommutators. Such a supertrace is not uniquely defined; it can always at least be modified by multiplication by an element of A.
Қарапайым із негізге тәуелді емес, сондықтан Z2 деңгейлі жағдайда қолдануға тиісті із – супертрек. Супертрек барлық T1, T2 ∈ End(V) үшін келесі қасиетті қанағаттандырады:
In the theory of superalgebras, if A is a commutative superalgebra, V is a free right A supermodule and T is an endomorphism from V to itself, then the supertrace of T, str(T) is defined by the following trace diagram:
More concretely, if we write out T in block matrix form after the decomposition into even and odd subspaces as follows,
then the supertrace
str(T) = the ordinary trace of T00 − the ordinary trace of T11. Let us show that the supertrace does not depend on a basis. Suppose e1, , ep are the even basis vectors and ep+1, , ep+q are the odd basis vectors. Then, the components of T, which are elements of A, are defined as
The grading of Tij is the sum of the gradings of T, ei, ej mod 2. A change of basis to e1', , ep', e(p+1)', , e(p+q)' is given by the supermatrix
and the inverse supermatrix
where of course, AA−1 = A−1A = 1 (the identity). We can now check explicitly that the supertrace is basis independent. In the case where T is even, we have
In the case where T is odd, we have
The ordinary trace is not basis independent, so the appropriate trace to use in the Z2 graded setting is the supertrace. The supertrace satisfies the property
for all T1, T2 in End(V). In particular, the supertrace of a supercommutator is zero. In fact, one can define a supertrace more generally for any associative superalgebra E over a commutative superalgebra A as a linear map tr: E > A which vanishes on supercommutators. Such a supertrace is not uniquely defined; it can always at least be modified by multiplication by an element of A.
More concretely, if we write out T in block matrix form after the decomposition into even and odd subspaces as follows,
then the supertrace
str(T) = the ordinary trace of T00 − the ordinary trace of T11. Let us show that the supertrace does not depend on a basis. Suppose e1, , ep are the even basis vectors and ep+1, , ep+q are the odd basis vectors. Then, the components of T, which are elements of A, are defined as
The grading of Tij is the sum of the gradings of T, ei, ej mod 2. A change of basis to e1', , ep', e(p+1)', , e(p+q)' is given by the supermatrix
and the inverse supermatrix
where of course, AA−1 = A−1A = 1 (the identity). We can now check explicitly that the supertrace is basis independent. In the case where T is even, we have
In the case where T is odd, we have
The ordinary trace is not basis independent, so the appropriate trace to use in the Z2 graded setting is the supertrace. The supertrace satisfies the property
for all T1, T2 in End(V). In particular, the supertrace of a supercommutator is zero. In fact, one can define a supertrace more generally for any associative superalgebra E over a commutative superalgebra A as a linear map tr: E > A which vanishes on supercommutators. Such a supertrace is not uniquely defined; it can always at least be modified by multiplication by an element of A.
str(T1T2) = str(T2T1). Әсіресе, суперкоммутатордың супертрегі нөлге тең. Шын мәнінде, кез келген ассоциативтік супералгебра E үшін коммутативтік супералгебра A арқылы анықталған сызықтық карта tr: E → A ретінде супертректі жалпылауға болады, ол суперкоммутаторларда нөлге тең болады. Мұндай супертрек бірегей анықталмайды; оны әрқашан кем дегенде A элементімен көбейту арқылы өзгертуге болады.
In the theory of superalgebras, if A is a commutative superalgebra, V is a free right A supermodule and T is an endomorphism from V to itself, then the supertrace of T, str(T) is defined by the following trace diagram:
More concretely, if we write out T in block matrix form after the decomposition into even and odd subspaces as follows,
then the supertrace
str(T) = the ordinary trace of T00 − the ordinary trace of T11. Let us show that the supertrace does not depend on a basis. Suppose e1, , ep are the even basis vectors and ep+1, , ep+q are the odd basis vectors. Then, the components of T, which are elements of A, are defined as
The grading of Tij is the sum of the gradings of T, ei, ej mod 2. A change of basis to e1', , ep', e(p+1)', , e(p+q)' is given by the supermatrix
and the inverse supermatrix
where of course, AA−1 = A−1A = 1 (the identity). We can now check explicitly that the supertrace is basis independent. In the case where T is even, we have
In the case where T is odd, we have
The ordinary trace is not basis independent, so the appropriate trace to use in the Z2 graded setting is the supertrace. The supertrace satisfies the property
for all T1, T2 in End(V). In particular, the supertrace of a supercommutator is zero. In fact, one can define a supertrace more generally for any associative superalgebra E over a commutative superalgebra A as a linear map tr: E > A which vanishes on supercommutators. Such a supertrace is not uniquely defined; it can always at least be modified by multiplication by an element of A.
More concretely, if we write out T in block matrix form after the decomposition into even and odd subspaces as follows,
then the supertrace
str(T) = the ordinary trace of T00 − the ordinary trace of T11. Let us show that the supertrace does not depend on a basis. Suppose e1, , ep are the even basis vectors and ep+1, , ep+q are the odd basis vectors. Then, the components of T, which are elements of A, are defined as
The grading of Tij is the sum of the gradings of T, ei, ej mod 2. A change of basis to e1', , ep', e(p+1)', , e(p+q)' is given by the supermatrix
and the inverse supermatrix
where of course, AA−1 = A−1A = 1 (the identity). We can now check explicitly that the supertrace is basis independent. In the case where T is even, we have
In the case where T is odd, we have
The ordinary trace is not basis independent, so the appropriate trace to use in the Z2 graded setting is the supertrace. The supertrace satisfies the property
for all T1, T2 in End(V). In particular, the supertrace of a supercommutator is zero. In fact, one can define a supertrace more generally for any associative superalgebra E over a commutative superalgebra A as a linear map tr: E > A which vanishes on supercommutators. Such a supertrace is not uniquely defined; it can always at least be modified by multiplication by an element of A.
Физиканың қолдануы
Суперсимметриялық кванттық өріс теорияларында, онда әрекет интегралы симметриялық түрлендірулер жиынтығы (суперсимметриялық түрлендірулер деп аталады) бойынша инвариант болып табылады, ал олардың алгебралары супералгебралар болып табылады, супертректің түрлі қолданыстары бар. Мұндай жағдайда, теорияның массалық матрицасының супертрегін әртүрлі спинді бөлшектердің массалық матрицаларының іздерінің спиндер бойынша қосындысы ретінде жазуға болады:
In supersymmetric quantum field theories, in which the action integral is invariant under a set of symmetry transformations (known as supersymmetry transformations) whose algebras are superalgebras, the supertrace has a variety of applications. In such a context, the supertrace of the mass matrix for the theory can be written as a sum over spins of the traces of the mass matrices for particles of different spin:
Аномалиясыз теорияларда, егер суперпотенциалда тек қайта нормаланатын мүшелер ғана болса, жоғарыда көрсетілген супертрек жоғалып кететінін көрсетуге болады, тіпті суперсимметрия спонтанды түрде бұзылған жағдайда да. Бір циклдық (кейде Колеман-Вайнберг потенциалы деп аталады) тиімді потенциалға үлестің супертректер арқылы да жазылуы мүмкін. Егер берілген теорияның массалық матрицасы болса, онда бір циклдық потенциалды былай жазуға болады:
In anomaly free theories where only renormalizable terms appear in the superpotential, the above supertrace can be shown to vanish, even when supersymmetry is spontaneously broken. The contribution to the effective potential arising at one loop (sometimes referred to as the Coleman–Weinberg potential) can also be written in terms of a supertrace. If is the mass matrix for a given theory, the one loop potential can be written as
мұнда және теориядағы бозондық және фермиондық еркіндік дәрежелері үшін тиісінше ағаш деңгейіндегі массалық матрицалар, ал – кесу шкаласы.
where and are the respective tree level mass matrices for the separate bosonic and fermionic degrees of freedom in the theory and is a cutoff scale.