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A Lie algebra representation is faithful when its representing homomorphism is injective.

A nonzero Lie algebra representation is irreducible when it has no proper nonzero invariant subspace.

For a finite-dimensional Lie algebra representation $\rho$ on $V$, the Trace form of a Lie algebra representation isWrite again $A=\rho (a)$, $B=\rho (b)$, and $C=\rho (c)=[A,B]$. The operator $C$ commutes with both $A$ and $B$. Direct use of the cyclic property of the trace givesIn the last line, cyclicity and $AC=CA$ turn $\mathrm{tr}(ABC)$ into $\mathrm{tr}(BAC)$. Thus the nonzero vector $c\in \mathfrak{g}$ is orthogonal to the basis $a,b,c$, and hence to all of $\mathfrak{g}$. The bilinear form is therefore degenerate.

Solved by gpt-5.6-sol high.

Use the Polynomial representation of the Heisenberg Lie algebra on the infinite-dimensional polynomial ring $\mathbb{C}[x]$:The product rule gives $[d/dx,x]=1$, so this is a Lie algebra representation. It is a Faithful Lie algebra representation: if $\alpha (d/dx)+\beta x+\gamma$ is the zero operator, applying it first to $1$ gives $\beta x+\gamma =0$, and then applying the remaining operator to $x$ gives $\alpha =0$.

To prove irreducibility, let $W$ be a nonzero invariant subspace and choose a nonzero polynomial in $W$ of least degree. If its degree were positive, repeated differentiation would produce a nonzero element of smaller degree, so $W$ contains a nonzero constant. Invariance under multiplication by $x$ then puts every monomial $x^n$ in $W$, and hence $W=\mathbb{C}[x]$.

Solved by gpt-5.6-sol high.

The Adjoint representation of a Lie algebra isIt is linear. The Jacobi identity givesso it is a Lie algebra representation.

Solved by gpt-5.6-sol high.

For a finite-dimensional Lie algebra representation $\rho :\mathfrak{g}\to \mathfrak{gl}(V)$, its trace form is the symmetric bilinear formIt is invariant: $([x,y],z{)}_V=(x,[y,z]{)}_V$.