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Mar 19, 2019 at 23:01 comment added Jules Lamers I haven't read that, but would expect that $\mathfrak{g}_{\mathbb{C}} = \mathfrak{g} \otimes_{\mathbb{R}} \mathbb{C}$ is the complexification of $\mathfrak{g}$ (see also the notation for $\mathfrak{p}_c$) --- and that $\mathfrak{g}$ would therefore probably be defined over the reals.
Mar 19, 2019 at 23:00 review Close votes
Mar 21, 2019 at 17:02
Mar 19, 2019 at 22:34 history asked Johnny T. CC BY-SA 4.0