What is the maximal number of orthogonal operators $ A _{1} , \dotsc, A _ {m} $ in $ \mathbb{Q}^{ n } $ satisfying the relations $ A_{i}^{2} = - I $ and $ A_{i}A_{j} + A_{j}A_{i} = 0 $ for $ i \neq j $ ? We can consider $ n= 6 $ .

If we take $ \mathbb{R} ^{n} $ instead of $ \mathbb{Q} ^{n} $ then the answer of the Hurtwitz–Radon problem is $ \rho(n) - 1 $ where if $ n = (2a + 1 ) 2^{c+4d} $ where $ 0 \leq c \leq 3 $ then $ \rho(n) = 2^{c} + 8d $.

What is the maximal number of orthogonal operators $ A _{1} , \dotsc, A _ {m} $ in $ \mathbb{Q}^{ n } $ satisfying the relations $ A_{i}^{2} = - I $ and $ A_{i}A_{j} + A_{j}A_{i} = 0 $ for $ i \neq j $ ? We can consider $ n= 6 $ .

If we take $ \mathbb{R} ^{n} $ instead of $ \mathbb{Q} ^{n} $ then the answer of the Hurwitz–Radon problem is $ \rho(n) - 1 $ where if $ n = (2a + 1 ) 2^{c+4d} $ where $ 0 \leq c \leq 3 $ then $ \rho(n) = 2^{c} + 8d $.

$($ Hurtwitz Radon $)$ what is the What is the maximal number of orthogonal operators $ A _{1} , . . . , A _ {m} $ in $ \mathbb{Q}^{ n } $ satisfying the relations $ A_{i}^{2} = - I $ and $ A_{i}A_{j} + A_{j}A_{i} = 0 $ for $ i \neq j $ ? we can consider $ n= 6 $ .

If we take $ \mathbb{R} ^{n} $ instead of $ \mathbb{Q} ^{n} $ then the answer of the Hurtwitz Radon problem is $ \rho(n) - 1 $ where if $ n = (2a + 1 ) 2^{c+4d} $ where $ 0 \leq c \leq 3 $ then $ \rho(n) = 2^{c} + 8d $

What is the maximal number of orthogonal operators $ A _{1} , \dotsc, A _ {m} $ in $ \mathbb{Q}^{ n } $ satisfying the relations $ A_{i}^{2} = - I $ and $ A_{i}A_{j} + A_{j}A_{i} = 0 $ for $ i \neq j $ ? We can consider $ n= 6 $ .

If we take $ \mathbb{R} ^{n} $ instead of $ \mathbb{Q} ^{n} $ then the answer of the Hurtwitz–Radon problem is $ \rho(n) - 1 $ where if $ n = (2a + 1 ) 2^{c+4d} $ where $ 0 \leq c \leq 3 $ then $ \rho(n) = 2^{c} + 8d $.