Hurwitz 1, 2, 4, 8 squared Theorem

We are all familiar with multiplication of plural numbers: If $ Z_1 = x_1 + iy_1, Z_2 = X_2 + iy_2 $ is two plural numbers, then $ | z_1z_2 | = | Z_1 | \ cdot | z_2 | $.

\ [(X_1 ^ 2 + Y_1 ^ 2) (X_2 ^ 2 + Y_2 ^ 2) = (x_1x_2-y_1y_2) ^ 2 + (x_1y_2 + x_2y_1) ^ 2. \]

In 1748, Euler discovered the following four squares equation:

\ [(X_1 ^ 2 + x_2 ^ 2 + X_3 ^ 2 + x_4 ^ 2) (y_1 ^ 2 + Y_2 ^ 2 + y_3 ^ 2 + y_4 ^ 2) = Z_1 ^ 2 + Z_2 ^ 2 + Z_3 ^ 2 + Z_4 ^ 2. \]

Where

\ [\ Left \ {\ ign in {Align *} & Z_1 = x_1y_1-x_2y_2-x_3y_3-x_4y_4, \ & Z_2 = x_1y_2 + x_2y_1 + x_3y_4-x_4y_3, \ & Z_3 = x_1y_3 + x_3y_1-x_2y_4 + x_4y_2, \ & Z_4 = x_1y_4 + x_4y_1 + x_2y_3-x_3y_2. \ end {Align *} \ right. \]

4. The sum of squares equation indicates that the norm in the Hamilton ry is still multiplier. In 1848, caley found the eight yuan number and exported a similar eight-square-sum equation. Of course, it will be very complicated to write it out. Let's just press it here.

Generally, if the multiplication between vectors can be defined on $ N $ dimension European space $ \ mathbb {r} ^ N $: \ [\ mathbb {R ^ n} \ times \ mathbb {R ^ n} \ rightarrow \ mathbb {R ^ n} :( V, W) \ rightarrow V \ times W. \] makes $ V \ times W $ linear for two components $ V and W $, and the product's norm is equal to the product of the norm: $ | V \ times w | = | v | \ cdot | w | $ (here $ | \ cdot | $ is a common European norm ), then we get an equation of the sum of squares of $ N $.

Over the next 50 years, people have been trying to find a possible 16-square-sum equation, but all failed, so they began to wonder if such an equation was true. Finally, Hurwitz proved the conclusion in 1898:

[Hurwitz sum of squares theorem]Set $ x = (x_1, \ ldots, X_n) $, $ Y = (y_1, \ dots, Y_n) $ to the vector in $ \ mathbb {r} ^ N $. If the bilinear functions $ Z_1 (x, y), \ ldots, Z_N (x, y) $ for $ X and Y $ make the equation

\ [(X_1 ^ 2 + \ cdots + x_n ^ 2) (y_1 ^ 2 + \ cdots + Y_n ^ 2) = Z_1 ^ 2 + \ cdots + Z_N ^ 2 \]

Then $ n = 1, 2, 4, 8 $.

As mentioned above, huiwitz's sum of squares theorem indicates $ \ mathbb {r }$ in the real number field, $ \ mathbb {c} $ in the complex number field, element (Euclidean) norm and vector multiplication are compatible in element $ \ mathbb {H} $ and octal number $ \ mathbb {o} $, in other dimensions of $ \ mathbb {r} ^ N $, vector multiplication compatible with European norm cannot be defined.

Hurwitz's proof is purely linear algebra. The proof of linear algebra is relatively Elementary, but the steps are slightly longer. In 1943, Eckmann gave a beautiful proof by using the finite group theory. However, this article uses linear algebra.

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Proof of Hurwitz Theorem

The first step to solve the problem is to convert the problem into a matrix equation. Set $ z = (Z_1, \ ldots, Z_N) $, then $ Z $ about $ y $ is linear, therefore, $ N $ level matrix $ A $ satisfies $ z = ya $. Of course, matrix $ A $ is related to $ x $. Therefore, the equation in Hurwitz's theorem becomes \ [(X_1 ^ 2 + x_2 ^ 2 + \ cdots + x_n ^ 2) YY '= Yaa 'y '. \] Because $ y $ is an indefinite element, \ [AA '= (X_1 ^ 2 + \ cdots + x_n ^ 2) I _n. \]

Furthermore, because $ A $ is linear about $ x $, set $ A = a_1x_1 + \ cdots + a_nx_n $.

\ [AA '= \ sum _ {I = 1} ^ na_ia_ I 'x _ I ^ 2 + \ sum _ {1 \ Leq I <J \ Leq n} (a_ia_j' + a_ja_ I ') x_ix_j. \]

Then we get a matrix equation.

\ [A_ia_ I '= I _n, \ quad a_ia_j' + a_ja_ I '= 0 \ quad \ Text {for} \ I \ ne J. \]

You can further normalize $ a_n $ into a matrix of units: make $ q_ I = a_ia_n ^ {-1} $, so $ Q_1, \ ldots, Q _ {n-1} $ meets

\ [Q_ I '=-q_ I, \ quad q_ I ^ 2 =-I _n, \ quad q_iq_j =-q_jq_ I \ quad \ Text {for} \ I \ ne J. \]

Obviously, $ N $ must be an even number (all the negative matrix terminologies of the odd order are 0), while $ n = 2 $ makes the conclusion true, therefore, we assume that $ n> 2 $, so the possible value of $ N $ is $, 8, \ cdots. $

Next, this theorem is an independent conclusion. You can forget the previous derivation to consider a new question: Set $ r$ to an even number, $ x_1, \ ldots, x_r $ is $ r$ real reversible matrix of $ N $ order, two-to-one inverse exchange: $ x_ix_j + x_jx_ I = 0 $ when $ I \ ne J $. Consider the matrix with the following shapes:

\ [S = \ left \ {M \ | \ m = X_1 ^ {E_1} X_2 ^ {E_2} \ cdots x_r ^ {e_r }, \ e_ I = 0, 1 \ right \}. \]

And define $ X_1 ^ 0x_2 ^ 0 \ cdots x_r ^ 0 = I _n $. This matrix has a total of $2 ^ r$. We need to prove it.

Theorem 1:All $2 ^ r$ matrices in $ S $ are linearly independent, resulting in $2 ^ r \ Leq n ^ 2 $.

$ R$ in the condition is an even number, which seems to be a strange restriction, but it is used in proof.

Proof of the theorem: the inverse proof method is used: first, the inverse exchange between $ X_ I $ is not difficult to see for any $ I $ and $ m \ in S $ x_im = \ PM mx_ I $.

Suppose there is a linear relationship between matrices in $ S $ \ [a_1m_1 + a_2m_2 + \ cdots + a_km_k = 0. \ Quad (\ AST) \] Obviously $ k \ geq 2 $. We can set this linear relationship to be the smallest length in all possible linear relationships (I .e., any matrix with less than $ K $ in $ S $ is linearly independent ), thus, each coefficient $ a_ I $ is not 0. For $ X_ I $, if $ X_ I $ is exchanged with $ m_j $, for example, $ M_1 and $ x_im_1 $ are exchanged: $ x_im_1 = m_1x_ I $, $ (\ AST) $ use $ X_ I $ for the combination of the two sides:

\ [A_1x_im_1x_ I ^ {-1} + a_2x_im_2x_ I ^ {-1} + \ cdots + a_kx_im_kx_ I ^ {-1} = 0, \]

\ [A_1m_1 + A_2'm _ 2 + \ cdots + A_k 'M _ k = 0. \]

Here, each $ a_j '= \ PM a_j (2 \ Leq J \ Leq K) $ is determined by the plus and minus signs. If $ x_im_j = m_jx_ I $, $ a_j' = a_j $, otherwise, $ a_j' =-a_j $. Subtract this equation from $ (\ AST) $ to get

\ [(A_2-a_2 ') M_2 + \ cdots + (a_k-a_k') m_k = 0. \]

In this way, we obtain a linear relationship with a length less than $ K $. Therefore, each $ J $ must have $ a_j = a_j' $, therefore, $ x_im_j = m_jx_ I $ is available for each $ m_j $, that is, $ X_ I $. As long as it is exchanged with $ M_1 $, it must be exchanged with $ M_1, \ ldots, and m_k $, in this way, we can find a very critical information.

Each $ X_ I $ is exchanged with $ M_1, \ ldots, and m_k $, or with $ M_1, \ ldots, and m_k $.

Now, for $ M = X_1 ^ {E_1} X_2 ^ {E_2} \ cdots x_r ^ {e_r} \ in S $, define the vector $ E (m) = (e_1, \ ldots, e_r) $, here each $ e_ I = 0, 1 $. Defines the vector $ \ delta (m) = (\ delta_1, \ ldots, \ delta_r) $, where $ x_im = mx_ I $ is $ \ delta_ I = 1 $, otherwise, $ \ delta_ I =-1 $. According to our analysis in the previous section, $ \ delta (M_1) = \ cdots = \ delta (m_k) $. Next we will explain that the $ \ delta (m) $ vector can uniquely determine $ E (m) $, and thus $ E (M_1) = \ cdots = E (m_k) $, that is, $ M_1 = \ cdots = m_k $. Of course, this is in conflict with $ M_1, \ cdots, and m_k $, therefore, the $2 ^ r$ matrices in $ S $ are linearly independent.

Set $ W (m) $ to the number of 1 in the $ E (m) $ component, then $ \ delta_ I = (-1) ^ {W (m) + e_ I} $. This is obtained by case-by-case analysis: If $ W (m) $ is an even number, the $ X_ I $ exchanged with $ M $ is exactly those that do not appear in the $ M $ expression (that is, those that correspond to $ e_ I = 0 $ ); if $ W (m) $ is an odd number, the opposite is true, the $ X_ I $ exchanged with $ M $ is exactly the ones that appear in the $ M $ expression (corresponding to the $ e_ I = 1 $ section $ X_ I $ ).

So $ \ delta_1 \ delta_2 \ cdots \ delta_r = (-1) ^ {r \ cdot W (m) + W (m)} = (-1) ^ {W (m)} $, where $ r$ is an even number. Therefore

\ [(-1) ^ {e_ I} = \ delta_ I (\ delta_1 \ delta_2 \ cdots \ delta_r). \]

Because $ e_ I $ only takes two values, 1, and 1, we can use $ \ delta (m) $ to calculate the value of each $ e_ I $, which proves the theorem.

Finally, let's return to the proof of Hurwitz's theorem. Recall the conclusion we have obtained: we have found the $ n-1 $ pairs of reverse swapping matrices $ B _1, \ ldots, B _ {n-1} $, and each $ B _ I =-B _ I '$, $ B _l ^ 2 =-I _n $. We know that $ N $ must be an even number and $ n-1 $ is an odd number. In order to apply the theorem, we need to remove a matrix $ B _ {n-1} $, matrix $ B _1, \ ldots, B _ {N-2} $ to satisfy the criteria of the theorem, so $2 ^ {N-2} \ Leq n ^ 2 $, and thus $ n =, 8 $.

There is also a situation where $ n = 6 $ that needs to be ruled out. To this end, we only need to prove the following supplementary conclusions:

Theorem 2: Set two linear transformations $ A and B $ in the $ N $ dimension European space to meet the requirements of \ [A =-A', B =-B ', A ^ 2 = B ^ 2 =-I _n, AB =-ba, \] Then $ N $ is a multiple of 4.

Proof: For any non-zero vector $ V $ in the space, it is easy to verify that the four vectors $ V, AV, BV, and ABV $ are orthogonal to each other, so they are linearly independent, thus, they form a 4-dimensional sub-space $ W $, and $ W $ is a constant sub-space of $ \ {A, B \} $. Consider the orthogonal complement space of $ W $ w ^ \ bot $, then $ w ^ \ bot $ is also the constant subspaces of $ A and B $, therefore, we can find a 4-dimensional Constant sub-space $ W_1 \ subset w ^ \ bot $, in this way, the whole space is divided into the straight sum of several four-dimensional subspaces, so the dimension $ N $ of the whole space is a multiple of 4.

Hurwitz 1, 2, 4, 8 squared Theorem