# Revision history [back]

### Where does this minus come from in Clifford with a positive quadratic form?

In the Clifford algebra squared, I find a minus I cannot explain:

C.<a0,a1>=CliffordAlgebra(DiagonalQuadraticForm(ZZ,[1,1])) ; C.rename('Cl(Z,2)')
a0.tensor(a0) in C.tensor(C)
a0.tensor(a1) * a0.tensor(a1)
# result: -1#1    ( why? )


In mathematics $a_0\otimes a_1$ squared equals its components squared, so $a_0^2\otimes a_1^2$ which should be the unit of the Cl(Z,2) above. Am I missing something?

### Where does this minus come from in Clifford with a positive quadratic form?

In the Clifford algebra squared, I find a minus I cannot explain:

C.<a0,a1>=CliffordAlgebra(DiagonalQuadraticForm(ZZ,[1,1])) ; C.rename('Cl(Z,2)')
a0.tensor(a0) in C.tensor(C)
a0.tensor(a1) * a0.tensor(a1)
# result: -1#1    ( why? )


In mathematics $a_0\otimes a_1$ squared equals its components squared, so $a_0^2\otimes a_1^2$ which should be the unit of the Cl(Z,2) above. Am I missing something?

I supposed first that Sage interprets the wandering of the, say, left $a_1$ to the 'right of the second' $a_0$ in

$(a_0\otimes a_1) \cdot (a_0\otimes a_1)$

as some anticommutation rule, but this is not what happens in mathematics (if I'm not mistaken) and renaming by creating a copy of the Clifford algebra with new-names for generators doesn't work.

### Where does this minus come from in Clifford with a positive quadratic form?

In the Clifford algebra squared, tensor-squared, I find a minus I cannot explain:

C.<a0,a1>=CliffordAlgebra(DiagonalQuadraticForm(ZZ,[1,1])) ; C.rename('Cl(Z,2)')
a0.tensor(a0) in C.tensor(C)
a0.tensor(a1) * a0.tensor(a1)
# result: -1#1    ( why? )


In mathematics $a_0\otimes a_1$ squared equals its components squared, so $a_0^2\otimes a_1^2$ which should be the unit of the Cl(Z,2) above. Am I missing something?

I supposed first that Sage interprets the wandering of the, say, left $a_1$ to the 'right of the second' $a_0$ in

$(a_0\otimes a_1) \cdot (a_0\otimes a_1)$

as some anticommutation rule, but this is not what happens in mathematics (if I'm not mistaken) and renaming by creating a copy of the Clifford algebra with new-names for generators doesn't work.

### Where does this minus come from in Clifford with a positive quadratic form?

In the Clifford algebra tensor-squared, I find a minus I cannot explain:

C.<a0,a1>=CliffordAlgebra(DiagonalQuadraticForm(ZZ,[1,1])) ; C.rename('Cl(Z,2)')
a0.tensor(a0) in C.tensor(C)
a0.tensor(a1) * a0.tensor(a1)
# result: -1#1    ( why? )


In mathematics $a_0\otimes a_1$ squared equals its components squared, so $a_0^2\otimes a_1^2$ which should be the unit of the Cl(Z,2) above. Am I missing something?

I supposed first that Sage interprets the wandering of the, say, left $a_1$ to the 'right of the second' $a_0$ in

$(a_0\otimes a_1) \cdot (a_0\otimes a_1)$

as some anticommutation rule, but this is not what happens in mathematics (if I'm not mistaken) and renaming by creating a copy of the Clifford algebra with new-names for generators doesn't work.

### Where does this minus come from in Clifford with a positive quadratic form?

In the Clifford algebra tensor-squared, I find a minus I cannot explain:

C.<a0,a1>=CliffordAlgebra(DiagonalQuadraticForm(ZZ,[1,1])) ; C.rename('Cl(Z,2)')
a0.tensor(a0) in C.tensor(C)
a0.tensor(a1) * a0.tensor(a1)
# result: -1#1    ( why? )


In mathematics $a_0\otimes a_1$ squared equals its components squared, so $a_0^2\otimes a_1^2$ which should be the unit of the Cl(Z,2) above. Am I missing something?

I supposed first that Sage interprets the wandering of the, say, left $a_1$ to the 'right of the second' $a_0$ in

$(a_0\otimes a_1) \cdot (a_0\otimes a_1)$

as some anticommutation rule, but this is not what happens in mathematics (if I'm not mistaken) and renaming by creating a copy of the Clifford algebra with new-names for generators doesn't work.

### Where does this minus come from in Clifford Cl$(\mathbb Z,2)^{\otimes 2}$ with a positive quadratic form?

In the Clifford algebra tensor-squared, I find a minus I cannot explain:

C.<a0,a1>=CliffordAlgebra(DiagonalQuadraticForm(ZZ,[1,1])) ; C.rename('Cl(Z,2)')
a0.tensor(a0) in C.tensor(C)
a0.tensor(a1) * a0.tensor(a1)
# result: -1#1    ( why? )


In mathematics $a_0\otimes a_1$ squared equals its components squared, so $a_0^2\otimes a_1^2$ which should be the unit of the Cl(Z,2) above. Am I missing something?

I supposed first that Sage interprets the wandering of the, say, left $a_1$ to the 'right of the second' $a_0$ in

$(a_0\otimes a_1) \cdot (a_0\otimes a_1)$

as some anticommutation rule, but this is not what happens in mathematics (if I'm not mistaken) and renaming by creating a copy of the Clifford algebra with new-names for generators doesn't work.

### Where does this minus come from in Cl$(\mathbb Z,2)^{\otimes Z,\mathrm{diag}[1,1])^{\otimes 2}$ with a positive quadratic form??

In the Clifford algebra tensor-squared, I find a minus I cannot explain:

C.<a0,a1>=CliffordAlgebra(DiagonalQuadraticForm(ZZ,[1,1])) ; C.rename('Cl(Z,2)')
a0.tensor(a0) in C.tensor(C)
a0.tensor(a1) * a0.tensor(a1)
# result: -1#1    ( why? )


In mathematics $a_0\otimes a_1$ squared equals its components squared, so $a_0^2\otimes a_1^2$ which should be the unit of the Cl(Z,2) above. Am I missing something?

I supposed first that Sage interprets the wandering of the, say, left $a_1$ to the 'right of the second' $a_0$ in

$(a_0\otimes a_1) \cdot (a_0\otimes a_1)$

as some anticommutation rule, but this is not what happens in mathematics (if I'm not mistaken) and renaming by creating a copy of the Clifford algebra with new-names for generators doesn't work.

### Where does this minus come from in the squared Clifford algebra, Cl$(\mathbb Z,\mathrm{diag}[1,1])^{\otimes 2}$ ?

In the Clifford algebra tensor-squared, I find a minus I cannot explain:

C.<a0,a1>=CliffordAlgebra(DiagonalQuadraticForm(ZZ,[1,1])) ; C.rename('Cl(Z,2)')
a0.tensor(a0) in C.tensor(C)
a0.tensor(a1) * a0.tensor(a1)
# result: -1#1    ( why? )


In mathematics $a_0\otimes a_1$ squared equals its components squared, so $a_0^2\otimes a_1^2$ which should be the unit of the Cl(Z,2) $\mathrm{Cl}(\mathbb Z,2) \otimes \mathrm{Cl}(\mathbb Z,2)$ above. Am I missing something?

I supposed first that Sage interprets the wandering of the, say, left $a_1$ to the 'right of the second' $a_0$ in

$(a_0\otimes a_1) \cdot (a_0\otimes a_1)$

as some anticommutation rule, but this is not what happens in mathematics (if I'm not mistaken) and renaming by creating a copy of the Clifford algebra with new-names for generators doesn't work.

### Where does this minus come from in the squared Clifford algebra, Cl$(\mathbb Z,\mathrm{diag}[1,1])^{\otimes 2}$ ?

In the Clifford algebra tensor-squared, I find a minus I cannot explain:

C.<a0,a1>=CliffordAlgebra(DiagonalQuadraticForm(ZZ,[1,1])) ; C.rename('Cl(Z,2)')
a0.tensor(a0) in C.tensor(C)
a0.tensor(a1) * a0.tensor(a1)
# result: -1#1    ( why? )


In mathematics $a_0\otimes a_1$ squared equals its components squared, so $a_0^2\otimes a_1^2$ which should be the unit of the $\mathrm{Cl}(\mathbb Z,2) \otimes \mathrm{Cl}(\mathbb Z,2)$ above. Am I missing something?

I supposed first that Sage interprets the wandering of the, say, left $a_1$ to the 'right of the second' $a_0$ in

$(a_0\otimes a_1) \cdot (a_0\otimes a_1) = a_0^2\otimes a_1^2$

as some anticommutation rule, anti-commutation rule ($a_0a_1 + a_1 a_0 =0$, hence the minus), but this is not what happens such products never happened in mathematics the computation (if I'm not mistaken) and mistaken). I tried renaming by creating a copy of the Clifford algebra with new-names for generators generators, but that doesn't work.work either.

### Where does this minus come from in the squared Clifford algebra, Cl$(\mathbb Z,\mathrm{diag}[1,1])^{\otimes 2}$ ?

In the Clifford algebra tensor-squared, I find a minus I cannot explain:

C.<a0,a1>=CliffordAlgebra(DiagonalQuadraticForm(ZZ,[1,1])) ; C.rename('Cl(Z,2)')
a0.tensor(a0) in C.tensor(C)
a0.tensor(a1) * a0.tensor(a1)
# result: -1#1    ( why? )


In mathematics $a_0\otimes a_1$ squared equals its components squared, so $a_0^2\otimes a_1^2$ which should be the unit of the $\mathrm{Cl}(\mathbb Z,2) \otimes \mathrm{Cl}(\mathbb Z,2)$ above. Am I missing something?

I supposed first that Sage interprets the wandering of the, say, left $a_1$ to the 'right of the second' $a_0$ in

$(a_0\otimes a_1) \cdot (a_0\otimes a_1) = a_0^2\otimes a_1^2$

as some anti-commutation rule ($a_0a_1 + a_1 a_0 =0$, hence the minus), but such products never happened in the computation (if (i.e. they must not occur for this example, if I'm not mistaken). I tried renaming creating a copy of the Clifford algebra with new-names for generators, but that doesn't work either.

### Where does this minus come from in the squared Clifford algebra, Cl$(\mathbb Z,\mathrm{diag}[1,1])^{\otimes 2}$ ?

In the Clifford algebra tensor-squared, I find a minus I cannot explain:

C.<a0,a1>=CliffordAlgebra(DiagonalQuadraticForm(ZZ,[1,1])) ; C.rename('Cl(Z,2)')
a0.tensor(a0) in C.tensor(C)
a0.tensor(a1) * a0.tensor(a1)
# result: -1#1    ( why? )


In mathematics $a_0\otimes a_1$ squared equals its components squared, so $a_0^2\otimes a_1^2$ which should be the unit of the $\mathrm{Cl}(\mathbb Z,2) \otimes \mathrm{Cl}(\mathbb Z,2)$ above. Am I missing something?

I supposed first that Sage interprets the wandering of the, say, left $a_1$ to the 'right of the second' $a_0$ in

$(a_0\otimes a_1) \cdot (a_0\otimes a_1) = a_0^2\otimes a_1^2$

as some anti-commutation rule ($a_0a_1 + a_1 a_0 =0$, hence the minus), but such products never happened in the computation (i.e. they must not occur for this example, if I'm not mistaken). I tried renaming creating a copy of the Clifford algebra with new-names for generators, but that doesn't work either.

### Where does this minus come from in the squared Clifford algebra, Cl$(\mathbb Z,\mathrm{diag}[1,1])^{\otimes 2}$ ?

In the Clifford algebra tensor-squared, I find a minus I cannot explain:

C.<a0,a1>=CliffordAlgebra(DiagonalQuadraticForm(ZZ,[1,1])) ; C.rename('Cl(Z,2)')
a0.tensor(a0) in C.tensor(C)
a0.tensor(a1) * a0.tensor(a1)
# result: -1#1    ( why? )


In mathematics $a_0\otimes a_1$ squared equals its components squared, so $a_0^2\otimes a_1^2$ which should be the unit of the $\mathrm{Cl}(\mathbb Z,2) \otimes \mathrm{Cl}(\mathbb Z,2)$ above. Am I missing something?

I supposed first that Sage interprets the wandering of the, say, left $a_1$ to the 'right of the second' $a_0$ in

$(a_0\otimes a_1) \cdot (a_0\otimes a_1) = a_0^2\otimes a_1^2$= 1\otimes 1$(trivial quad. form) as some anti-commutation rule ($a_0a_1 + a_1 a_0 =0\$, hence the minus), but such products never happened in the computation (i.e. they must not occur for this example, if I'm not mistaken). I tried renaming creating a copy of the Clifford algebra with new-names for generators, but that doesn't work either.