The Killing form is defined by $$\newcommand{\ad}{\operatorname{ad}} K(u,v) = \operatorname{tr} (\ad_u \circ \ad_v).$$
Here tr is the trace in the sense of linear algebra, which is canonically defined, and $\ad$ is the adjoint representation of the Lie algebra on itself. In the case of a matrix Lie algebra, since the adjoint representation is given by $$\ad_u (w) = [u, w]$$ and the bracket is the matrix commutator, you can work out that the Killing form actually agrees with the trace formula, at least up to a scale factor. This is to be expected because the Killing form is with conditions, the, up to scale, unique $\operatorname{Ad}$-invariant non-degenerate quadratic form, and the trace formula has this invariance too.
The condition for this uniqueness is that $G$ should be compact and simple. If $G$ is semi-simple, there is one choice of scale per factor, $a\kappa_1 \oplus b\kappa_2$ is a $\operatorname{Ad}$-invariant on $\mathfrak{g}_1 \oplus \mathfrak{g}_2$ if $\kappa_1, \kappa_2$ are on $\mathfrak{g}_1, \mathfrak{g}_2$. Also for the case of $G = U(1)$, there is a unique, up to scale, $\kappa$: linearly map $\mathfrak{u}(1) \to i\mathbb{R}$ in any way (i.e. choose any basis vector) and define $\kappa$ through multiplication of real numbers. It is a theorem that for a direct sum of Lie algebras where each term is either compact and simple or $\mathfrak{u}(1)$ the symmetric, $\operatorname{Ad}$-invariant bilinear exists and is positive definite. (I learned about this and read the proof in Weinberg's The Quantum Theory of Fields, vol II, but I'm sure it's covered in a Lie algebra textbook.)
On Wikipedia you find listed the Killing form for some common Lie algebras. In particular in every case there except $\mathfrak{gl}(n)$ the listed Killing form is proportional to the trace. $GL(n)$ is of course not compact, but it is covered by the previous paragraph. You can see that when $X,Y$ are traceless, and belong to $\mathfrak{sl}(n)\subset\mathfrak{gl}(n)$, we recover the Killing form for the former.
I assume that we are dealing with Lie algebras over an field $K$. Let me try to give an overview about how this induced representation is constructed, which is an exercise in understanding how representations of the Lie algebra $\mathfrak{g}$ relate to modules of the universal enveloping algebra $\mathcal{U}(\mathfrak{g})$. A book which you might want to look into is James E. Humphreys’ Introduction to Lie Algebras and Representation Theory, where the topic is covered from an algebraic point of view and without the use of Lie groups.
The universal enveloping algebra
As you probably already know that the universal eneveloping algebra $\mathcal{U}(\mathfrak{g})$ of the Lie algebra $\mathfrak{g}$ is an associative, untial $K$-algebra that can be defined as the quotient algebra $T(\mathfrak{g})/I$ where $T(\mathfrak{g})$ is the tensor algebra over $\mathfrak{g}$ and $I$ the two-sided ideal
$$
I = \langle x \otimes y - y \otimes x - [x,y] \mid x,y \in \mathfrak{g}\rangle \,.
$$
So as a vector space, $\mathcal{U}(\mathfrak{g})$ is generated by the monomials
$$
x_1 \dotsm x_n
\qquad
\text{with $x_1, \dotsc, x_n \in \mathfrak{g}$}.
$$
The inclusion map $\mathfrak{g} \hookrightarrow \mathcal{U}(\mathfrak{g})$ is a homomorphism of Lie algebras, so we can regard $\mathfrak{g}$ as a Lie subalgebra of $\mathcal{U}(\mathfrak{g})$. The multiplication in $\mathcal{U}(\mathfrak{g})$ satisfies the property
$$
xy-yx = [x,y]_{\mathfrak{g}}
$$
for all $x,y \in \mathfrak{g} \subseteq \mathcal{U}(\mathfrak{g})$.
An essential property that follows from this construction of $\mathcal{U}(\mathfrak{g})$ is the theorem of Poincaré–Birkhoff–Witt, which decribes a vector space basis of $\mathcal{U}(\mathfrak{g})$.
Theorem (Poincaré–Birkhoff–Witt). Let $\mathfrak{g}$ be a Lie algebra and let $(x_i)_{i \in I}$ be a basis of $\mathfrak{g}$ where $(I, \leq)$ is a totally ordered set. Then the ordered monomials
$$
x_{i_1} \dotsm x_{i_n}
\qquad
\text{with $n \in \mathbb{N}$, $i_1, \dotsc, i_n \in I$, $i_1 \leq \dotsb \leq i_n$}
$$
form a basis of $\mathcal{U}(\mathfrak{g})$.
Remark: This basis can also be written as
$$
x_{i_1}^{p_1} \dotsm x_{i_m}^{p_m}
$$
with $m \in \mathbb{N}$, $i_1, \dotsc, i_n \in I$, $i_1 < \dotsb < i_n$, $p_1, \dotsc, p_n \in \mathbb{N}$.
Let’s look at a specific example:
Example: The Lie algebra $\mathfrak{sl}_2(\mathbb{C})$ admits a basis $(e,h,f)$ given by
$$
e =
\begin{pmatrix}
0 & 1 \\
0 & 0
\end{pmatrix},
\quad
h =
\begin{pmatrix}
1 & 0 \\
0 & -1
\end{pmatrix},
\quad
f =
\begin{pmatrix}
0 & 0 \\
1 & 0
\end{pmatrix}.
$$
This basis is already totally ordered by the order in which we write the tupel $(e,h,f)$. By the PBW-theorem it follows that $\mathcal{U}(\mathfrak{sl}_2(\mathbb{C}))$ has a basis consisting of the the monomials
$$
e^l h^m f^n \quad \text{with $l,m,n \in \mathbb{N}$.}
$$
One nice thing about the universal enveloping algebra is its universal property:
Proposition (Universal property of the UEA).
Let $\mathfrak{g}$ be a $K$-Lie algebra and let $A$ be an associative and unital $K$-algebra.
Any Lie algebra homomorphism $\phi \colon \mathfrak{g} \to A$ extends uniquely to an algebra homomorphism $\Phi \colon \mathcal{U}(\mathfrak{g}) \to A$.
Representations of $\mathfrak{g}$ and $\mathcal{U}(\mathfrak{g})$-modules
This universal property has nice effects for the representation theory of $\mathfrak{g}$.
A representation of $\mathfrak{g}$ is, by definition, a Lie algebra homomorphism $\mathfrak{g} \to \mathfrak{gl}(V)$ for some vector space $V$.
Using the universal property of $\mathcal{U}(\mathfrak{g})$ it follows that the Lie algebra homomorphisms $\mathfrak{g} \to \mathfrak{gl}(V)$ correspond one-to-one to the algebra homomorphisms $\mathcal{U}(\mathfrak{g}) \to \mathrm{End}_K(V)$ by restricting or extending the homomorphisms in question.
But an algebra homomorphism from $\mathcal{U}(\mathfrak{g})$ to $\mathrm{End}_K(V)$ is the same as a $\mathcal{U}(\mathfrak{g})$-module structure on $V$.
We have thus found a one-to-one correspondence between representations of $\mathfrak{g}$ and $\mathcal{U}(\mathfrak{g})$-modules.
Let us be more explicit:
If $\rho \colon \mathfrak{g} \to \mathfrak{gl}(V)$ is a representation of $\mathfrak{g}$, then we have a multiplication map (i.e. a bilinear map)
$$
\mathfrak{g} \times V \to V,
\quad
(x,v) \mapsto x.v
$$
given by
$$
x.v = \rho(x)(v)
$$
for all $x \in \mathfrak{g}$ and $v \in V$.
This multiplication now extends to a multiplication map
$$
\mathcal{U}(\mathfrak{g}) \times V \to V,
\quad
(y,v) \mapsto y \cdot v
$$
that is given with respect to the PBW-basis of $\mathcal{U}(\mathfrak{g})$ by
$$
(x_1 \dotsm x_n) \cdot v
=
x_1 . (x_2 . ( \dotsm ( x_{n-1} . (x_n.v) ) ) )
$$
for all $x_1, \dotsc, x_n \in \mathfrak{g}$ and $v \in V$.
This multiplication gives $V$ the structure of an $\mathcal{U}(\mathfrak{g})$-module.
Note also that to the restriction of this $\mathcal{U}(\mathfrak{g})$-module structure to $\mathfrak{g}$ (when regarded as a subset of $\mathcal{U}(\mathfrak{g})$ coincides with the representation of $\mathfrak{g}$ that we started with.
Extension of scalars
We will now need the tensor product, and in particular its use for the so called extension of scalars. A detailed explanation can, for example, be found in Abstract Algebra by Dummit and Foote.
The basic idea behind the induced representation will be precisely this extension of scalars.
Let $R$ is a unitial ring, let $S$ be a subring of $R$ (also unital) and let $M$ be a (unital) $S$-module.
Then we want to somehow extend the $S$-module structure of $M$ to an $R$-module structure on $M$.
The problem is that there is a priori no good way to extends the multiplication map $S \times M \to M$ to a multiplication map $R \times M \to M$.
So instead, we replace $M$ by another abelian group, namely $R \otimes_S M$.
Recall that $R \otimes_S M$ is an abelian group generated by elements of the form $r \otimes m$ with $r \in R$ and $m \in M$, under the constraints that
$$
(r+r') \otimes m = r \otimes m + r' \otimes m, \\
r \otimes (m+m') = r \otimes m + r \otimes m', \\
(rs) \otimes m = r \otimes (sm)
$$
where $r, r' \in R$, $m, m' \in M$ and $s \in S$.
The nice thing about $R \otimes_S M$ is that it naturally carries the structure of an $R$-module via
$$
r' \cdot (r \otimes m)
= (r' r) \otimes m
$$
for all $r', r \in R$ and $m \in M$.
Note that for all $s \in S$ and $m \in M$ we have
$$
s \cdot (1 \otimes m)
= (s \cdot 1) \otimes m
= s \otimes m
= (1 \cdot s) \otimes m
= 1 \otimes (sm),
$$
so we can, to some extend, think of $R \otimes_S M$ as an extending the original $S$-module structure of $M$. Indeed, the map
$$
M \to R \otimes_S M \,,
\quad
m \mapsto 1 \otimes m
$$
is a homomorphism of $S$-modules. (A word of warning though: It is not always true that this homomorphism is injective. So we can not necessarily regard $M$ as an $S$-submodule of $R \otimes_S M$.)
We refer to this process of “extending” the $S$-module structure of $M$ to the $R$-module structure on $R \otimes_S M$ as the extension of scalars from $S$ to $R$.
The induced representation
We are now ready to tackle the induced representation. For this let $\mathfrak{g}$ be a Lie algebra, let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$ a Lie subalgebra and let $\rho \colon \mathfrak{h} \to \mathfrak{gl}(V)$ a representation of $\mathfrak{h}$. We then have for all $x \in \mathfrak{h}$ and $v \in V$ the product $x.v = \rho(x)(v)$.
Note that the inclusion map $\mathfrak{h} \hookrightarrow \mathfrak{g}$ is a homomorphism of Lie algebras and therefore induces (by the universal property of the universal enveloping algebra) a homomorphism of algebras $\mathcal{U}(\mathfrak{h}) \to \mathcal{U}(\mathfrak{g})$ that is given on the PBW-bases of $\mathcal{U}(\mathfrak{h})$ and $\mathcal{U}(\mathfrak{g})$ by
$$
x_1 \dotsm x_n \mapsto x_1 \dotsm x_n
\quad
\text{for all $x_1, \dotsc, x_n \in \mathfrak{h}$}.
$$
This homomorphism of algebras is injective since it maps the PBW-basis of $\mathcal{U}(\mathfrak{h})$ injectively into the PBW-basis of $\mathcal{U}(\mathfrak{g})$.
We can therefore regard $\mathcal{U}(\mathfrak{h})$ as a subalgebra of $\mathcal{U}(\mathfrak{g})$.
We can now apply the extension of scalars from $\mathcal{U}(\mathfrak{h})$ to $\mathcal{U}(\mathfrak{g})$:
That $V$ is a representation of $\mathfrak{h}$ means that it is a $\mathcal{U}(\mathfrak{h})$-module via
$$
(x_1 \dotsm x_n) \cdot v
=
x_1.(x_2.(\dotsm(x_{n-1}.(x_n.v))))
$$
for all $x_1, \dotsc, x_n \in \mathfrak{h}$ and $v \in V$.
So by applying extension of scalars we get the $\mathcal{U}(\mathfrak{g})$-module
$$
\mathrm{Ind}_\mathfrak{h}^\mathfrak{g}(V)
=
\mathcal{U}(\mathfrak{g}) \otimes_{\mathcal{U}(\mathfrak{h})} V.
$$
But let’s see how the $\mathcal{U}(\mathfrak{g})$-module structure on $\mathrm{Ind}_\mathfrak{h}^\mathfrak{g}(V)$ really looks like:
- As a vector space, $\mathrm{Ind}_\mathfrak{h}^\mathfrak{g}(V)$ is generated by the simple tensors $x \otimes v$ with $x \in \mathcal{U}(\mathfrak{g})$ and $v \in V$, and $\mathcal{U}(\mathfrak{g})$ is generated by the monomial $x_1 \dotsm x_n$ with $x_1, \dotsc, x_n \in \mathfrak{g}$.
It follows that $\mathrm{Ind}_\mathfrak{h}^\mathfrak{g}(V)$ is, again as a vector space, generated by the elements
$$
(x_1 \dotsm x_n) \otimes v
\quad
\text{with $x_1, \dotsc, x_n \in \mathfrak{g}$, $v \in V$}.
$$
In terms of these vector space generators, the $\mathcal{U}(\mathfrak{g})$-module structure of $\mathrm{Ind}_\mathfrak{h}^\mathfrak{g}(V)$ is given by the multiplication
$$
(x_1 \dotsm x_n) \cdot ((y_1 \dotsm y_m) \otimes v)
= (x_1 \dotsm x_n \cdot y_1 \dotsm y_m) \otimes v.
$$
- The fact that we are tensoring over $\mathcal{U}(\mathfrak{h})$ has the effect that
$$
h \cdot (1 \otimes v)
= h \otimes v
= 1 \otimes (h.v)
\quad
\text{for all $h \in \mathfrak{h}$, $v \in V$}.
$$
These two properties can be nicely put together by choosing a suitable basis of $\mathfrak{g}$ and applying the PBW theorem: Let $(b_i)_{i \in I}$ be a basis of $\mathfrak{g}$ where
- $(I ,\leq)$ is a totally ordered set,
- we have a partition $I = J' \cup J$ so that $(b_j)_{j \in J}$ is a basis of $\mathfrak{h}$, and
- $j' \leq j$ for all $j' \in J'$ and $j \in J$.
Then by the PBW-theorem, the ordered monmials
$$
b_{i_1} \dotsm b_{i_n} b_{j_1} \dotsm b_{j_m}
\qquad
\begin{alignedat}{2}
i_1, \dotsc, i_n &\in J', \,
&
i_1 \leq \dotsb &\leq i_n,
\\
j_1, \dotsc, j_m &\in J, \,
&
j_1 \leq \dotsb &\leq _m
\end{alignedat}
$$
form a basis of $\mathcal{U}(\mathfrak{g})$, and the ordered monomials
$$
b_{j_1} \dotsm b_{j_m}
\qquad
\text{with $j_1, \dotsc, j_m \in J$, $j_1 \leq \dotsb \leq j_m$}
$$
form a basis of the subalgebra $\mathcal{U}(\mathfrak{h})$.
With respect to this basis, the $\mathcal{U}(\mathfrak{g})$-module structure on $\mathrm{Ind}_\mathfrak{h}^\mathfrak{g}(V)$ is given by
$$
(b_{i_1} \dotsm b_{i_n} b_{j_1} \dotsm b_{j_m}) . (1 \otimes v)
= (b_{i_1} \dotsm b_{i_n}) \otimes (b_{j_1}.(\dotsm (b_{j_m}.v)))
$$
(Roughly speaking: The basis elements of $\mathfrak{h}$ just do what they have always done, and the “new” basis elements are put in front. Just as expected from a formal and general construction.)
Example:
Let $\mathfrak{g} = \mathfrak{gl}_2(\mathbb{C})$ and $\mathfrak{h} = \mathfrak{sl}_2(\mathbb{C})$, and let $(e,h,f)$ be the basis of $\mathfrak{sl}_2(\mathbb{C})$ as in the previous example. We can extend this basis of $\mathfrak{sl}_2(\mathbb{C})$ to a basis $(b,e,h,f)$ of $\mathfrak{gl}_2(\mathbb{C})$ by choosing
$$
b =
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}.
$$
This basis $(b, e, h, f)$ of $\mathfrak{gl}_2(\mathbb{C})$ is already totally ordered by the way its elements are ordered in the tuple $(b, e, h, f)$. A basis of $\mathcal{U}(\mathfrak{gl}_n(\mathbb{C}))$ is now given by the monomials
$$
b^k e^l h^m f^n
\quad
\text{with $k,l,m,n \in \mathbb{N}$}
$$
and a basis of $\mathcal{U}(\mathfrak{sl}_2(\mathbb{C}))$ is given by the monomials
$$
e^l h^m f^n
\text{with $l,m,n \in \mathbb{N}$}.
$$
Consider now the natural representation $V = \mathbb{C}^2$ of $\mathfrak{sl}_2(\mathbb{C})$, which is given by
$$
x.v
=
xv
\text{for all $x \in \mathfrak{sl}_2(\mathbb{C})$ and $v \in \mathbb{C}^2$},
$$
where the right hand side denotes the usual matrix-vector-multiplication.
Then the induced $\mathcal{U}(\mathfrak{gl}_2(\mathbb{C}))$-module structure on $\mathrm{Ind}_\mathfrak{h}^\mathfrak{g}(V)$ is given by
$$
(b^k e^l h^m f^n) \cdot (1 \otimes v)
= b^k \otimes (e^l h^m f^n v)
\quad
\text{for all $v \in \mathbb{C}^2$}.
$$
Note that we we do not get the natural representation of $\mathfrak{gl}_2(\mathbb{C})$, as $b$ does not act by the identity, but rather via $b.(1 \otimes v) = b \otimes v$ for all $v \in \mathbb{C}^2$.
We are now nearly finished: We have constructed a $\mathcal{U}(\mathfrak{g})$-module structure on $\mathrm{Ind}_\mathfrak{h}^\mathfrak{g}(V)$, which by restriction to $\mathfrak{g}$ (when regarded as a subset of $\mathcal{U}(\mathfrak{g})$) now correspond to a representation of $\mathfrak{g}$. This representation is given by
$$
x . ((x_1 \dotsm x_n) \otimes v)
= (x \cdot x_1 \dotsm x_n) \otimes v
\quad
\text{for all $x, x_1, \dotsc, x_n \in \mathfrak{g}$, $v \in V$},
$$
a special case of the previous formulae.
I hope this helps.
Best Answer
$\DeclareMathOperator\g{\mathfrak{g}}\DeclareMathOperator\h{\mathfrak{h}}$The first preliminary remark is that $\mathfrak{z}(\g)\cap [\g,\g]$ is in the kernel of $B_\rho$ for every rep $\rho$. Indeed, for this we can pass to an algebraic closure, block-trigonalize the representation with irreducible diagonal blocks: on each (irreducible) diagonal block, this central intersection acts by scalars of trace zero, hence by zero.
So, combined with Ado's theorem, the assumption ensures that $\mathfrak{z}(\g)\cap [\g,\g]=0$. Se can write $\g$ as $\mathfrak{z}(\g)\times\h$. By Ado's theorem choose a faithful rep for $\h$, and take the direct sum with a faithful rep for the central factor so that it acts by strict upper triangular matrices. The direct sum of those reps is a faithful rep of the direct product, with degenerate associated bilinear form unless $\mathfrak{z}(\g)=0$.
Hence, $\g$ has trivial center, and this case is already mentioned by the OP, using the adjoint representation.
(Note that the result trivially implies Ado's theorem, so we can't expect to bypass it.)