Why are superalgebras so vital?

I recognize that a superalgebra is a $\mathbb Z/2\mathbb Z$-graded algebra which it acts perfectly. I recognize really little physics however, so despite the fact that I recognize that the extremely- prefix relates to supersymmetry, I do not recognize what that suggests; exists an engaging mathematical factor to take into consideration superalgebras?

2019-05-06 01:58:36
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Answers: 3

I can sum up one actually standard factor, which is in fact the factor I initially obtained curious about the definition. Take a limited - dimensional vector room $V$ of measurement $n$, and also allow $\left( {n \choose k} \right) = {n+k-1 \choose k}$ represent the variety of multisets of dimension $k$ on a set of dimension $n$. (Multisets resemble parts other than that greater than one duplicate of an offered component is feasible.) After that the symmetrical powers of $V$ have measurements

$\displaystyle \left( {n \choose 1} \right), \left( {n \choose 2} \right), ... $

whereas the outside powers of $V$ have measurements

$\displaystyle {n \choose 1}, {n \choose 2}, ....$

Now below is an amusing identification : it is not tough to see that ${n \choose k} = (-1)^k \left( {-n \choose k} \right)$. One means we could analyze this identification is that the $k^{th}$ outside power of a vector room of measurement $n$ resembles the $k^{th}$ symmetrical power of a vector room of measurement $-n$, whatever that suggests. So what could that perhaps suggest?

The solution (and also I'll allow you function this out on your own, due to the fact that it's enjoyable) is to operate in the group of supervector rooms! A supervector room $V$ is a straight amount $V_0 \oplus V_1$, and also while one idea of measurement is to take $\dim V_0 + \dim V_1$, an additional (which can be encouraged by thinking about $\mathbb{Z}/2\mathbb{Z}$ - rated vector rooms as the group of depictions of $\mathbb{Z}/2\mathbb{Z}$) is to take $\dim V_0 - \dim V_1$. So while a totally also vector room has favorable measurement, a totally weird vector room has adverse measurement.

After that : rated - commutativity indicates that the symmetrical power of a totally weird vector room is the outside power of the equivalent totally also vector room. Extra usually, the $k^{th}$ symmetrical power of a vector room of measurement $n$ (for all integers $n$) has measurement $\left( {n \choose k} \right)$.

(And, certainly, the symmetrical algebra of a supervector room is normally a rated - commutative superalgebra.)

(The physics link is that symmetrical powers = bosons, outside powers = fermions, and also there is a duality in between both.)

2019-05-08 13:59:16

Sure, supermodules over a superalgebra create a wonderful and also nontrivial instance of a tensor group. Actually, it is a theory that Deligne that all tensor groups over $\mathbb{C}$ with ideal development problems can be gotten as groups of extremely - depictions.

An additional factor is offered by Qiaochu, that supercommutativity emerges normally in a great deal of areas.
As an example, the outside algebra is supercommutative ; hence the wedge item on differential kinds acts similarly. Surprisingly, the outside algebra of a straight amount represents taking the extremely tensor item.

An additional instance : With single cohomology on a topological room, there is a means to specify a mug item, which please the supercommutative regulation $ab = (-1)^{\deg b \deg a} ba$.

(Note that offered a $\mathbb{Z}$ - rated extremely - commutative algebra, you can get a $\mathbb{Z}/2$ - rated superalgebra as you show in your inquiry by taking the amount of the weird components and also the amount of the even components.)

2019-05-08 09:39:52

If you agree to construct a formalism of supergeometry after that some building and constructions are less complicated to state in this language, especially differential kinds and also the De Rham facility. The first one is simply features from the weird line ($R^{(0 | 1)}$) to your manifold, and also the De Rham differential originates from extremely - diffeomorphisms acting upon that line.

Additionally, there is a symplectic/orthogonal duality in depiction concept that some individuals (most notoriously Kontsevich) have actually supported as ideal recognized in regards to Lie superalgebras.

Constructions making use of an officially adverse - dimensional object (or estimations that resemble they could originate from such an object) occasionally can be analyzed in regards to authentic $Z/2$ rated things whose superdimension (also measurement minus weird measurement) is the adverse measurement concerned.

modify : you will certainly get more educated solutions if you upload to Mathoverflow, where several of the factors have actually created documents on extremely - or noncommutative geometry.

2019-05-08 09:29:49