Geometric Algebra Space-Time ❡

Format ❡

Basis Axes ❡

γ₀ is time and squares to -1

γ₁ is space x, squares to 1
γ₂ is space y, squares to 1
γ₃ is space z, squares to 1

So that's Cl(3,1). THE ONLY TRUE DEFINITION. jk, some seem to do Cl(1,3) but ill stay with what is taught.

Bivectors ❡

Now the planes are

γ₂₃ all space thus ² = -1
γ₃₁ all space thus ² = -1
γ₁₂ all space thus ² = -1
γ₀₁ time and space ² = +1
γ₀₂ time and space ² = +1
γ₀₃ time and space ² = +1

unsure why?

Lets see γ₂₃² = γ₂₃₂₃ = -γ₂₂₃₃ so -1(γ₂² = -1 γ₃² = -1) + -1 * -1 * -1 = -1

Lets see γ₀₁² = γ₀₁₀₁ = -γ₀₀₁₁ so -1(γ₀² = +1 γ₁² = -1) + -1 * +1 * -1 = +1

No shame, i was certain it's still -1 once .... but it's 1.

Trivectors ❡

When we form trivectors here we will NOT get 4 positive, 4 negative ones like in 4,0 4D.

there are 4 3 2 * 1 = 24 possible permutations of 4 bases. BUT we only want 3 of them. So thats 4! / 3! = 4 basis trivectors.

Because all have positive and negative possible values thats 8 "cubes". But lets stay with the 4 basis trivectors.

Now ask to define them: What axis is left out? And there it makes sense to go 0,1,2,3 is left out

Then we check what happens when we wedge this trivector with the missing axis. We know (you can check below at Quadvector) that I² = -1 in STA.

γ₁: γ₀₂₃ ∧ γ₁ = γ₀₂₃₁ (2 swaps) = +γ₀₁₂₃ = +I (+I² = -1)
γ₂: γ₀₃₁ ∧ γ₂ = γ₀₁₃₂ (2 swaps) = +γ₀₁₂₃ = +I (+I² = -1)    <-- damn!
γ₃: γ₀₁₂ ∧ γ₃ = γ₀₁₂₃ (0 swaps) = +γ₀₁₂₃ = +I (+I² = -1)
γ₀: γ₁₂₃ ∧ γ₀ = γ₁₂₃₀ (3 swaps) = -γ₀₁₂₃ = -I (-I² = +1)

@TBH I THINK this has to do with all 8 cubes needing to be summed to a negative volume. To be checked!!

Quadvector ❡

in normal 4D we had I² = 1.

e₁₂₃₄² = e₁₂₃₄₁₂₃₄ = -e₁₂₃₁₄₂₃₄ = e₁₂₁₃₄₂₃₄ = -e₂₃₄₂₃₄ = e₂₃₂₄₃₄ = -e₃₄₃₄ = 1

NOT THE CASE IN STA!!!

γ₀₁₂₃ is I and I² = -1!

I² = γ₀₁₂₃² = γ₀₁₂₃₀₁₂₃ = -γ₀₁₂₀₃₁₂₃ = γ₀₁₀₂₃₁₂₃ = -γ₀₀₁₂₃₁₂₃ = γ₀₀₁₂₁₃₂₃ = -γ₀₀₁₁₂₃₂₃ = γ₀₀₁₁₂₂₃₃ = 1(-1 1 1 1) = -1