I want to see that the fundamental representation is a representation. Suppose the structure constants $f^{abc}$ are given. We can assume there is at least one non-zero structure constant, otherwise any set of commuting $d \times d$ Hermitian matrices would comprise a representation. So assume that $a,b,c$ are such that $f^{abc} \neq 0$. According to Peskin & Schroeder (1995), the $k$-dimensional vector $\xi^n$ is the (1-dimensional) fundamental representation over some field $\mathbb{F}$. If so then $$ i f^{abc} \xi^c \stackrel{!}{=} [\xi^a,\xi^b] = \xi^a\xi^b - \xi^b\xi^a = \xi^a\xi^b - \xi^a\xi^b = 0 $$ since $\xi^n \in \mathbb{F}$ for each $n$, thus $\xi^a$ commutes with $\xi^b$. This is a contradiction since we assumed that there exists at least one $f^{abc} \neq 0$.

(1) What is wrong?

(2) How can we express $\xi^a$ in terms of $f^{abc}$, such that we see that the fundamental representation is indeed a representation?

EDIT: The pages in Peskin are 498-499. According to Peskin page 498 the definition of a d-dimensional representation is a set of $d \times d$ Hermitian matrices $\xi^a$ such that for given structure constants $f^{abc}$, $if^{abc}\xi^c = [\xi^a,\xi^b]$. According to Peskin page 499, in the fundamental representation of a Lie group of dimension $k$, there exists some field $\mathbb{F}$ such that the $k$-dimensional vector over $\mathbb{F}$ is the fundamental representation.

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    $\begingroup$ 1. What is your definition of "fundemental representation"? 2. What is your definition of "representation"? I'm asking because the question doesn't make any sense in the standard mathematical approach - the fundamental representation is a representation by its very definition, and it doesn't make sense to say that "the $k$-dimensional vector $\xi^n$ is the (1-dimensional) fundamental representation", nor why one could apply a Lie bracket to such a vector. Another data point for "don't learn representation theory from physicists", I guess. $\endgroup$ – ACuriousMind Nov 19 '17 at 19:24
  • $\begingroup$ Which page in P&S? What eq. #s? $\endgroup$ – Qmechanic Nov 19 '17 at 19:25
  • $\begingroup$ I added the pages in P&S as well as translated the crude definitions given in P&S to what he intends to say in terms of mathematics (at least as I understand it) $\endgroup$ – Mikkel Rev Nov 19 '17 at 19:34
  • $\begingroup$ @ACuriousMind It makes perfect sense to say that $k$-dimensional vector $\xi^n$ is a 1 dimensional representation, for if each component $\xi^n$ is self-adjoint, then the $n$-th component, $\xi^n$ is 1-dimensional Hermitian matrix with respect to the field $\mathbb{F}$. $\endgroup$ – Mikkel Rev Nov 19 '17 at 19:36
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    $\begingroup$ The post (v6) seems to be a misinterpretation of what P&S are saying. Consider to give exact block quotes. $\endgroup$ – Qmechanic Nov 19 '17 at 19:52

Actually your question is due a misunderstanding. For a representation 2 things are necessary. First of all, there are elements $\xi^a$, $\xi^b$ of a Lie-algebra (actually these are the generators of the Lie-algebra) which fulfill a commutator relation which indeed defined by the structure constants like

$[\xi^a, \xi^b]= f_c^{ab} \xi^c$

Summation is done over double indices. This is a abstract relation where is not said how it is realized. In order to see an realization of it, the concept of a representation is needed. But for (here finite) representation 2 concepts are necessary.

1) a vector space $V$ (k-dimensional if the representation is finite) is needed, and indeed the components of V could be of any field $\mathbb{F}$ (but typically $\mathbb{R}$ or $\mathbb{C}$).

2) a mapping which maps the elements of the Lie-algebra into elements into the space of endomorphisms $\mathfrak{gl}(V)$ which act on $V$. If the Lie-algebra is called $\mathfrak{gl}$, then the representation is a mapping from

$ \rho: \mathfrak{gl} \rightarrow \mathfrak{gl}(V)$.

So to make the math a bit more transparent, by the representation an abstract Lie-algebra element $\xi^a$ is mapped into a matrix $\Xi^a$ (those could be for instance hermitian, but this depends on the particular Lie-algebra) which acts then of vectors $v$ of the space $V$ so that $\Xi^a\cdot v$ remains in the space $V$.

Last but not least, the mapping $\rho$ must fulfill the (representation) property:

$\rho([\xi^a, \xi^b]) = \rho(\xi^a)\rho(\xi^b)-\rho(\xi^b)\rho(\xi^a) =f_{c}^{ab}\rho(\xi^c)\,\,\,$ (*)

For the fundamental representation, this property has to be checked. Of course this check should result for the fundamental representation that the property (*) is indeed fulfilled.

Finally what you are doing is:

$[\rho(\xi^a), \rho(\xi^b)]\cdot v = \rho(\xi^a)\cdot (\rho(\xi^b) \cdot v) - \rho(\xi^b)\cdot(\rho( \xi^a )\cdot v)$.

Clearly, in $\rho(\xi^a)$ are always matrices, never vectors, $v$ is the vector. And indeed for this generally k-dim. vector $k$ can be $k=1$. However, this representation operating on a 1-dim. vector is trivial, and definitely this is not the fundamental representation.

The purpose of the concept of representations is that the matrix operation defined by the Lie-algebra on $V$ stays inside $V$.

Finally, if you like you can "upgrade" (more details on wikipedia) the concept of a Lie-algebra representation to the corresponding Lie-group representations, where similar rules hold, in particular the representation property $\tilde{\rho}(g_1 g_2) =\tilde{\rho}(g_1)\tilde{\rho}(g_2)$ where $g_1$ and $g_2$ are 2 group members of group $\mathfrak{G}$ and

$\tilde{\rho}: g \in \mathfrak{G} \rightarrow \mathfrak{G}(V)$

where $\mathfrak{G}(V)$ are endomorphisms operating on $V$ fulfilling the representation property $\tilde{\rho}(g_1 g_2) =\tilde{\rho}(g_1)\tilde{\rho}(g_2)$, matrices which typically have different properties from those of the corresponding Lie-algebra. Further details have to be consulted in wikipedia which has an excellent documentation on representation theory of Lie-algebras and Lie-groups.

Finally, let's consider the fundamental representation of the Lie-algebra $\mathfrak{so}(3)$ which is characterized by the highest eigenvalue of its maximal abelian subalgebra which is for the fundamental one in physics notation $m=\frac{1}{2}$. Then the elements of the Lie-algebra are representated by the Pauli-matrices:

$[\sigma^a, \sigma^b] = \epsilon^{abc} \sigma^c$

The vectors $\psi$ the Pauli-matrices act on as a Lie-algebra representation of (2m+1)-dim. complex vectors are called non-relativistic spinors. And of course the mapping of the Lie-algebra $\mathfrak{so}(3)$ to all linear combinations of the Pauli-matrices fulfill the property (*). Of course many other "matrix" representations of the abstract Lie-algebra $\mathfrak{so}(3)$ are possible, with $v\in V$ 3-dim. or higher-dimensional. Probably $m=0$ is also possible (to be checked if the property $(*)$ is really fulfilled, but I guess, trivially it is), then you get the trivial representation acting on a (2m+1)=1-dim. vector. But that's not a fundamental representation.

  • $\begingroup$ Thanks for your response. So is the short answer that (i) the fundamental representation of SO(2) is the function $\rho$ that maps $\rho(\xi^c) = \sigma^c$, where $\sigma^c$ is Pauli matrix number $c$? (ii) And for the adjoint representation the map $\rho$ is the identity map? $\rho(\xi^c) = \xi^c$? $\endgroup$ – Mikkel Rev Nov 20 '17 at 15:12
  • $\begingroup$ 1) I spoke about $\mathfrak{so}(3)$, not about SO(2). 2) c runs from 1 to 3. 3) For the adjoint representation you have to define $\xi^c$ as an endomorphism on a (particular) vector space, otherwise $\xi^c$ simply remains an element in the abstract Lie-algebra, or differently said otherwise it cannot be a representation, i.e. you have to define what $\xi^c\cdot v$ is. $\endgroup$ – Frederic Thomas Nov 20 '17 at 21:04
  • $\begingroup$ Is it correct that $\rho$ is the representation? You explicitly write so in point (2). If this is true, what is the rule for $\rho : \mathfrak{gl} \to\mathfrak{gl}(V) $ in the fundamental representation? $\endgroup$ – Mikkel Rev Nov 20 '17 at 21:37
  • $\begingroup$ Again, you need 2 things: 1) the action on the vector space V (and this depends crucially on the type of vector space you consider), that's the endomorphism which is kind of parametrized by the elements of the abstract Lie-algebra. 2) Depending on your Lie-algebra element you get another endomorphism on V. This is the mapping $\rho$. $\endgroup$ – Frederic Thomas Nov 20 '17 at 21:46

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