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Oct 16, 2016 at 22:17 vote accept Hauke Reddmann
Oct 15, 2016 at 13:47 comment added Victor Protsak @Venky: Yes, with the additional assumptions you stated.
Oct 15, 2016 at 13:46 comment added Victor Protsak @YCor: Thank you, as a matter of fact, I was hoping the question could be reopened. Note that extra work is needed if the field is not algebraically closed or has char 2 (there exist even nondegenerate symmetric bilinear forms).
Oct 15, 2016 at 13:41 answer added Victor Protsak timeline score: 5
Oct 15, 2016 at 9:06 comment added Venkataramana I should add that the char zero field is algebraically closed
Oct 15, 2016 at 8:35 comment added Venkataramana @Protsak: does it mean that every semi-simple Lie algebra over char zero field has cyclic structure constants? (Because the Killing form is symmetric and non-degenerate)
Oct 15, 2016 at 8:00 history reopened YCor
Francois Ziegler
Loïc Teyssier
Dan Petersen
András Bátkai
Oct 15, 2016 at 7:02 comment added Dan Petersen In (unnecessarily) fancy language, such a Lie algebra is an algebra over the cyclic Lie operad.
Oct 15, 2016 at 5:37 review Reopen votes
Oct 15, 2016 at 8:00
Oct 15, 2016 at 5:20 comment added YCor I rewrote the question hoping that it will it be reopened so that Victor's answer can be posted.
Oct 15, 2016 at 5:17 history edited YCor CC BY-SA 3.0
Rewrote the question which was closed as unclear
Oct 14, 2016 at 20:55 comment added Victor Protsak Endow your Lie algebra with the symmetric bilinear form making $\{e_i\}$ an orthornormal basis (the Gram matrix is the identity matrix). Since $c_{ij}^k=([e_i,e_j],e_k)$, what you call "cyclicity" is equivalent to the invariance of the form.
Oct 14, 2016 at 19:02 history closed YCor
Wolfgang
Chris Godsil
Stefan Kohl
abx
Needs details or clarity
Oct 14, 2016 at 18:25 comment added Hauke Reddmann Typo. Sorry. Corrected.
Oct 14, 2016 at 18:24 history edited Hauke Reddmann CC BY-SA 3.0
Now where did the = come from?
Oct 13, 2016 at 5:25 review Close votes
Oct 14, 2016 at 19:05
Oct 12, 2016 at 20:55 comment added darij grinberg Why does $c^k_{ij} = c^i_{jk}$ define the semisimples? In $\mathfrak{sl}_2$, the bracket $\left[e,f\right]$ has a nonzero $h$-coordinate, but the bracket $\left[h,e\right]$ has no nonzero $f$-coordinate. Are you using a weirder basis?
Oct 12, 2016 at 20:46 history asked Hauke Reddmann CC BY-SA 3.0