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If V is a 3-dimensional Lie algebra with basis vectors E,F,G with Lie bracket relations [E,F]=G, [E,G]=0, [F,G]=0 and V' is the Lie algebra consisting of all 3x3 strictly upper triangular matrices with complex entries then would you say the following 2 mappings (isomorphisms) are different? I had to give an example of 2 different isomorphisms between these vector spaces.

φ:VV\varphi : V \to V' given by

φ(aE+bF+cG)=(0ac00b000)  ,          a,b,cC\varphi(aE+bF+cG)=\left( \begin{array}{ccc} 0 & a & c\\ 0 & 0 & b\\ 0 & 0 & 0 \end{array} \right)\;,\;\;\;\;\;a,b,c \in \mathbb{C}

and φˉ:VV\bar{\varphi} : V \to V' given by

φˉ(E)=(010000000)\bar{\varphi}(E)=\left( \begin{array}{ccc} 0 & 1 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{array} \right)

φˉ(F)=(000001000)\bar{\varphi}(F)=\left( \begin{array}{ccc} 0 & 0 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{array} \right)

φˉ(G)=(001000000)\bar{\varphi}(G)=\left( \begin{array}{ccc} 0 & 0 & 1\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{array} \right)
(edited 11 years ago)
Maybe this post is a little late given that the thread is a week old now, but recall that Lie algebra isomorphisms are also linear transformations, so are determined by their values on the basis elements E,F and G. So it seems your two maps are the same. However, the map you've given is an isomorphism and from this, it doesn't look too difficult to find a second?

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