Tensor product of quadratic forms

In <a href="/facts/Mathematics/pxTouaz4">mathematics</a>, the <a href="/facts/Tensor_product/7K3pR1ap">tensor product</a> of <a href="/facts/Quadratic_form/l2n1jXPe">quadratic forms</a> is most easily understood when one views the quadratic forms as <a href="/facts/Quadratic_space/l2n1jXPe">quadratic spaces</a>. If R is a <a href="/facts/Commutative_ring/QS1r2ovR">commutative ring</a> where 2 is <a href="/facts/Unit_(ring_theory)/GJSBu8Y7">invertible</a>, and if 
 
 
 
 (
 
 V
 
 1
 
 
 ,
 
 q
 
 1
 
 
 )
 
 
 {\displaystyle (V_{1},q_{1})}
 
 and 
 
 
 
 (
 
 V
 
 2
 
 
 ,
 
 q
 
 2
 
 
 )
 
 
 {\displaystyle (V_{2},q_{2})}
 
 are two quadratic spaces over R, then their tensor product 
 
 
 
 (
 
 V
 
 1
 
 
 ⊗
 
 V
 
 2
 
 
 ,
 
 q
 
 1
 
 
 ⊗
 
 q
 
 2
 
 
 )
 
 
 {\displaystyle (V_{1}\otimes V_{2},q_{1}\otimes q_{2})}
 
 is the quadratic space whose underlying R-<a href="/facts/Module_(mathematics)/jP0LIhFL">module</a> is the <a href="/facts/Tensor_product_of_modules/K3oaFIS4">tensor product</a> 
 
 
 
 
 V
 
 1
 
 
 ⊗
 
 V
 
 2
 
 
 
 
 {\displaystyle V_{1}\otimes V_{2}}
 
 of R-modules and whose quadratic form is the quadratic form associated to the tensor product of the <a href="/facts/Bilinear_form/gyvzn9ym">bilinear forms</a> associated to 
 
 
 
 
 q
 
 1
 
 
 
 
 {\displaystyle q_{1}}
 
 and 
 
 
 
 
 q
 
 2
 
 
 
 
 {\displaystyle q_{2}}
 
.
In particular, the form 
 
 
 
 
 q
 
 1
 
 
 ⊗
 
 q
 
 2
 
 
 
 
 {\displaystyle q_{1}\otimes q_{2}}
 
 satisfies

(
        
          q
          
            1
          
        
        ⊗
        
          q
          
            2
          
        
        )
        (
        
          v
          
            1
          
        
        ⊗
        
          v
          
            2
          
        
        )
        =
        
          q
          
            1
          
        
        (
        
          v
          
            1
          
        
        )
        
          q
          
            2
          
        
        (
        
          v
          
            2
          
        
        )
        
        ∀
        
          v
          
            1
          
        
        ∈
        
          V
          
            1
          
        
        ,
         
        
          v
          
            2
          
        
        ∈
        
          V
          
            2
          
        
      
    
    {\displaystyle (q_{1}\otimes q_{2})(v_{1}\otimes v_{2})=q_{1}(v_{1})q_{2}(v_{2})\quad \forall v_{1}\in V_{1},\ v_{2}\in V_{2}}

(which does uniquely characterize it however). It follows from this that if the quadratic forms are diagonalizable (which is always possible if 2 is invertible in R), i.e.,

q
          
            1
          
        
        ≅
        ⟨
        
          a
          
            1
          
        
        ,
        .
        .
        .
        ,
        
          a
          
            n
          
        
        ⟩
      
    
    {\displaystyle q_{1}\cong \langle a_{1},...,a_{n}\rangle }

q
          
            2
          
        
        ≅
        ⟨
        
          b
          
            1
          
        
        ,
        .
        .
        .
        ,
        
          b
          
            m
          
        
        ⟩
      
    
    {\displaystyle q_{2}\cong \langle b_{1},...,b_{m}\rangle }

then the tensor product has diagonalization

q
          
            1
          
        
        ⊗
        
          q
          
            2
          
        
        ≅
        ⟨
        
          a
          
            1
          
        
        
          b
          
            1
          
        
        ,
        
          a
          
            1
          
        
        
          b
          
            2
          
        
        ,
        .
        .
        .
        
          a
          
            1
          
        
        
          b
          
            m
          
        
        ,
        
          a
          
            2
          
        
        
          b
          
            1
          
        
        ,
        .
        .
        .
        ,
        
          a
          
            2
          
        
        
          b
          
            m
          
        
        ,
        .
        .
        .
        ,
        
          a
          
            n
          
        
        
          b
          
            1
          
        
        ,
        .
        .
        .
        
          a
          
            n
          
        
        
          b
          
            m
          
        
        ⟩
        .
      
    
    {\displaystyle q_{1}\otimes q_{2}\cong \langle a_{1}b_{1},a_{1}b_{2},...a_{1}b_{m},a_{2}b_{1},...,a_{2}b_{m},...,a_{n}b_{1},...a_{n}b_{m}\rangle .}

Tensor product of quadratic forms open-in-new

Tensor product of quadratic forms