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Jun 1 at 19:37 history edited LSpice CC BY-SA 4.0
Capitalise title, and name of paper, while this is on the front page
Feb 11, 2013 at 12:58 comment added Robert Bryant @Dmitry: I don't know a reference, but consider the element in $(\mathbb{C}^2)^{\otimes 3}$ given by $$ A_q = a_1\otimes\bigl(b_1\otimes c_1+b_2\otimes c_2\bigr) +a_2\otimes\bigl(b_1\otimes c_2-q\ b_2\otimes c_1\bigr) $$ where $q$ is a nonzero element of the field $\mathbb{F}$. $A_q$ and $A_p$ lie on the same orbit iff the quadratic forms $x^2+qy^2$ and $x^2+py^2$ are projectively equivalent over $\mathbb{F}$. When $\mathbb{F}=\mathbb{Q}$, there are an infinite number of such equivalent classes.
Feb 11, 2013 at 6:56 answer added Dan Fox timeline score: 6
Feb 10, 2013 at 20:09 comment added Dmitry Kerner @Robert: being ignorant I did not think that the field matters much. :( Could you give more details? A reference?
Feb 10, 2013 at 20:04 vote accept Dmitry Kerner
Feb 10, 2013 at 18:44 answer added Steven Sam timeline score: 8
Feb 10, 2013 at 17:06 comment added Robert Bryant @Dmitry: Also, could you be a bit more explicit about what you mean by 'discrete'? For example, in the case $n_1=n_2=n_3=2$, there are only a finite number of orbits when the ground field is $\mathbb{R}$ or $\mathbb{C}$, but there are infinitely many distinct orbits when the ground field is $\mathbb{Q}$. Would you count the latter case as 'discrete'?
Feb 10, 2013 at 14:51 comment added Robert Bryant Note that, because there is a $2$-parameter group of scalings in your equivalence group that acts trivially, you actually have that $$ n_1^2+n_2^2+n_3^2-2\ge n_1n_2n_3 $$ is the obvious necessary condition. (Thus, for example, $n_1=n_2=n_3=3$ must have invariants, though that wouldn't be obvious without subtracting the $2$.)
Feb 10, 2013 at 14:36 comment added Dima Pasechnik MathSciNet shows 2 references on the paper you cited, have a look!
Feb 10, 2013 at 14:34 comment added Dima Pasechnik In the case $n_1=n_2=n_3$, isn't it the same as classification of non-associative algebras? That was worked on a lot by A.Albert, etc.
Feb 10, 2013 at 13:50 history edited Dmitry Kerner CC BY-SA 3.0
added 100 characters in body
Feb 10, 2013 at 11:54 history asked Dmitry Kerner CC BY-SA 3.0