Killing form of isomorphic Lie algebras I tried to figure out how the killing form of two isomorphic finite dimensional Lie algebras over a field $k$ is related. I.e. if there is an isomorphism $\phi:K\to L$, how is $\kappa_K$ related to $\kappa_L$?
 A: Let me quickly expand Dietrich Burde's comment above.
Fix elements $X, Y \in K$. If we can show that $\kappa_L(\phi(X), \phi(Y)) = \kappa_K(X, Y)$ that will be enough for us, assuming that we don't use any special properties of $X$ and $Y$.
Let $n$ be the dimension of $K$. We fix a basis $X_1, X_2, \ldots, X_n$ of $K$ and define the $2n^2$ numbers $a_{i,j}$ and $b_{i,j}$ by $[X, X_i] = \sum_{j = 1}^n a_{i, j}X_j$ and $[Y, X_i] = \sum_{j = 1}^n b_{i, j}X_j$.
Let $A = (a_{i, j})$ and $B = (b_{i, j})$ be the corresponding $n \times n$ matrices. Anybody seeing just $A$ and $B$ would have no idea that there is a Lie algebra in the background (since they are completely ordinary matrices) but we know that $A$ represents $ad(X)$ and $B$ represents $ad(Y)$ w.r.t. to the basis $X_1, \ldots, X_n$.
In particular we have that 
$$\kappa_K(X, Y) = tr(AB).$$
Again: nothing on the right hand side refers visually to $K$.
So far so good.
Now since $\phi$ is an isomorphism we find that:
$$[\phi(X), \phi(X_i)] = \phi([X, X_i]) = \phi(\sum_j a_{i, j} X_j) = \sum_j a_{i, j} \phi(X_j)$$
The $a_{i,j}$ stayed the same throughout the above equations so we find that the same old matrix $A$ represents $ad(\phi(X))$ w.r.t. the basis $\phi(X_1), \ldots, \phi(X_n)$ of $L$.
Similar $B$ represents $ad(\phi(Y))$ and hence we find:
$$\kappa_L(\phi(X), \phi(Y)) = tr(AB).$$
Again, the left hand side comes from $L$ but on the right hand side we are just working with matrices and numbers. Now we can just cut the middle man and conclude that:
$$\kappa_K(X, Y) = \kappa_L(\phi(X), \phi(Y))$$
This answers your question. (Of course you can be distrustful of whether or not the answer would be different if we had used another basis than $X_1, \ldots, X_n$ but this is another issue about Killing forms in which the isomorphism $\phi$ plays no role.) 
