5.5 Gauge Group Hom($\cal {W}$, R₊)

We recall that the group $\it Hom$($\cal {W}$,$\bf R_{+}^{}$) is a set of homomorphisms from vector space $\cal {W}$ (treated as an abelian group of vectors) into group $\bf R_{+}^{}$. As there exists a natural isomorphism between the groups $\bf R_{+}^{}$ and $\bf R$:

exp : R $$\ni$$ y $$\longrightarrow$$ ey $$\in$$ $$\bf R_{+}^{}$$, ln : $$\bf R_{+}^{}$$ $$\ni$$ y $$\longrightarrow$$ lny $$\in$$ $$\bf R$$,

(5.14)

it suffices to describe the structure of group $\it Hom$($\cal {W}$,$\bf R$).

Let $\left\{\vphantom{ \widehat{e}_{j} }\right.$$\widehat{e}_{j}^{}$ $\left.\vphantom{ \widehat{e}_{j} }\right\}$(j = 1, 2,..., n) be the vector base for $\cal {W}$, then each element f $\in$ $\it Hom$($\cal {W}$, R) is described by a single line matrix [f₁,..., f_n]. The value f on the vector x $\in$ $\cal {W}$ is described in a natural manner:

$$\left(\vphantom{ \prod_{j=1}^{n} \widehat{e}_{j}^{x_{j}} }\right. \prod_{j=1}^{n} \widehat{e}_{j}^{x_{j}} \left.\vphantom{ \prod_{j=1}^{n} \widehat{e}_{j}^{x_{j}} }\right) \sum_{j=1}^{n}$$ (5.15)

Passing over from group R to $\bf R_{+}^{}$ from (5.9), (5.11) we obtain

$$\prod_{j=1}^{n}$$ (5.16)

Defining h_j = e^f_j we obtain the element h of group $\it Hom$($\cal {W}$,$\bf R_{+}^{}$) conformable to the matrix [f₁,..., f_n].:

$$\left(\vphantom{ \prod_{j=1}^{n} \widehat{e}_{j}^{x_{j}} }\right. \prod_{j=1}^{n} \widehat{e}_{j}^{x_{j}} \left.\vphantom{ \prod_{j=1}^{n} \widehat{e}_{j}^{x_{j}} }\right) \prod_{j=1}^{n}$$ (5.17)

In practice the homomorphism h $\in$ $\it Hom$($\cal {W}$,$\bf R_{+}^{}$) is most often described by giving the numbers h_j(j = 1, 2,..., n). The single-line matrix of homomorphism h in the base $\widehat{e}_{j}^{}$(j = 1, 2,..., n) is thus considered to be the ordered n-ple:

[h₁,..., h_n], (5.18)

where h_j $\in$ $\bf R_{+}^{}$ for j = 1, 2,..., n is obvious since (5.14), (5.15) are continuous mappings, the homomorphisms (5.17) are also continuous.

The here described procedure of choosing base $\left\{\vphantom{ \widehat{e}_{j} }\right.$$\widehat{e}_{j}^{}$ $\left.\vphantom{ \widehat{e}_{j} }\right\}$ and the parameterization of the group $\it Hom$($\cal {W}$,$\bf R_{+}^{}$) by matrices (5.18) are denoted as the choice of a dimensional base. This method has been in general use.

The gauge group does not depend either on the dimensional space construction. To describe $\it Hom$($\cal {W}$,$\bf R_{+}^{}$) it suffices to give the space of pure $\cal {W}$ dimensions. The gauge group is, therefore, a rather universal object.

Let r $\in$ $\it Hom$($\cal {W}$,$\bf R_{+}^{}$) be the fixed homomorphism of groups. The kernel of homomorphism r is defined and denoted:

Ker  r = {$$\omega$$ $$\in$$ $$\cal {W}$$ : r($$\omega$$) = 1}.

Of course, Ker  r is a subgroup of $\cal {W}$ as a kernel of the homomorphism of groups, moreover it is also a vector subspace.

Let the two different homomorphisms r₁, r₂ $\in$ $\it Hom$($\cal {W}$,$\bf R_{+}^{}$) have the same kernel Ker  r₁ = Ker  r₂. It is then easy to verify that one of them is the power of the other, i.e., there exists the number $\alpha$ $\in$ R for which r₂($\omega$) = r₁^$\alpha$($\omega$) (for $\omega$ $\in$ $\cal {W}$). Let $\omega$ $\in$ $\cal {W}$ = $\cal {W}$₀ $\oplus$ Ker  r₁, then there exists a unique decomposition of the vector

$$\omega$$ = $$\omega_{0}^{}$$v,

where $\omega_{0}^{}$ $\in$ $\cal {W}$₀, v $\in$ Ker  r₁. As dim$\cal {W}$₀ = 1 each vector $\omega$ has the representation

$$\omega$$ = $$\sigma^{a}_{0}$$v,

where $\sigma_{0}^{}$ is the fixed vector from $\cal {W}$₀, a $\in$ R and v $\in$ Ker  r₁. It is now easy to compare the values of the two homomorphisms:

$$\omega \sigma_{0}^{} \omega \sigma_{0}^{}$$ (5.19)

Since there always exists such a number $\alpha$ $\in$ R, that r₂($\sigma_{0}^{}$) = r₁^$\alpha$($\sigma_{0}^{}$), we obtain from (5.19)

$$\forall$$$$\omega$$ $$\in$$ $$\cal {W}$$        r₂($$\omega$$) = r₁^$\alpha$($$\omega$$).

Let us note at the same time that the equalities (5.19) are equivalent to condition

$$\forall \in \it Hom \cal {W} \bf R_{+}^{} \forall \omega \in \cal {W} \forall \in \omega^{a}_{} \omega$$ (5.20)

Thus (5.19) shows that the elements of the gauge group are in fact linear transformations and not merely group homomorphisms. Obtained results are presented as a theorem.

Theorem 5.1.   The gauge group $\it Hom$($\cal {W}$,$\bf R_{+}^{}$) is a vector space over R (in a multiplicative notation). The unit of space $\it Hom$($\cal {W}$,$\bf R_{+}^{}$) is the homomorphism e:

$$\forall$$$$\omega$$ $$\in$$ $$\cal {W}$$      e($$\omega$$) = 1,

the multiplication by a scalar are the powers:

$$\forall$$$$\alpha$$ $$\in$$ R    $$\forall$$r $$\in$$ $$\it Hom$$($$\cal {W}$$,$$\bf R_{+}^{}$$)        (r^$\alpha$)($$\omega$$) = r^$\alpha$($$\omega$$),

the vector product - the group product

$$\forall$$r₁, r₂ $$\in$$ $$\it Hom$$($$\cal {W}$$,$$\bf R_{+}^{}$$)    $$\forall$$$$\omega$$ $$\in$$ $$\cal {W}$$  (r₁r₂)($$\omega$$) = r₁($$\omega$$)r₂($$\omega$$).

The dimension of the vector space $\it Hom$($\cal {W}$,$\bf R_{+}^{}$) is identical with the dimension of $\cal {W}$ :

$$\it Hom \cal {W} \bf R_{+}^{} \cal {W}$$ (5.21)

Proof:
The fact that $\it Hom$($\cal {W}$,$\bf R_{+}^{}$) is a vector space results from (5.20) and the Definition 5.9. The equality of dimensions (5.21) results from the parameterization by matrices (5.18).

Finally, let us note that the gauge group is a rather non-typical object from the point of view of classical dimensional analysis. Let r $\in$ $\it Hom$($\cal {W}$,$\bf R_{+}^{}$), then r : $\cal {W}$ $\longrightarrow$ $\bf R_{+}^{}$ so that the arguments of homomorphism r are quatities of a dimensional type, e.g., mass, force, charge, while the values are numbers from $\bf R_{+}^{}$. Thus we have mappings of dimensions into dimensionless numbers.