Killing’s equation in a linear space

[LCSphysicist]

Homework Statement

I will post a print

Relevant Equations

N.

Be f a orthogonal transformation and g being in the canonical form.
$$[f^{-1}]^{t}[g][f^{-1}] = [g']$$
So this equation of isometries implies that the diagonal of g is +- 1, but, apparently, if it is minus one it can't be a group, and i don't know why.

To make the things clear, if det = +1, in the Lorentz metric the g is the Poincaré group

 

[jambaugh]

I am not clear on your question. Your statement about the diagonal of g, and "it" being -1 doesn't make sense to me. Can you state the original text of the problem you are trying to solve?

Also your stated equation seems to me to be mixing two distinct identities. For the matrix representation $[g]$ of a bilinear form $g$, its transformation under any linear transformation $f$ with matrix representation $[f]$ would be:

$$g\mapsto g' = [f^{-1}]^T [g][f^{-1}]$$
(or with $f\leftrightarrow f^{-1}$ reversed, depending on whether you think of the transformation as active or passive.)

That a linear transformation is orthogonal as defined by the metric means that it must leave the metric invariant and thus:
$$[f^{-1}]^T [g][f^{-1}]=g$$

The fact that the identity transformation is necessarily orthogonal and that composition of linear transformations is associative leads directly to the fact that all orthogonal transformations for a given metric forms a group. Call this group $O(V,g)$ where $V$ is the linear (vector) space on which $g$ is defined.

As to determinants, since matrix multiplication respects determinants (det of product is product of det's) you can make several statements about determinants of transformations. In particular:
$$det(g') = det(f^{-1})^2 det(g)$$
and thus for orthogonal transformations $det(f)^2 = +1$.

Review your question with the above in mind.

 

[LCSphysicist]

Hi, it is not a question, it is a doubt about the text i was reading, probably this is a justification to the confusion in the OP, i am still trying to get the theory :)

Anyway, i think you give the det of f, but i am asking for the det of g, to be clear, i will post a image of the text, sorry any misunderstanding, i am trying to get it yet ;)

Do you see the why on the text?That is actually the question, why the negative det do not form a group.

 

[LCSphysicist]

Ok, i believe i get the answer, if we define a group of det (-1), and their elements A and B, the product of their owns elements leave outside the group, that is: det(AB) = det(A)det(B) = 1, so it is not a group at all.

@fresh_42 that is it??

 

[fresh_42]

Ok, i believe i get the answer, if we define a group of det (-1), and their elements A and B, the product of their owns elements leave outside the group, that is: det(AB) = det(A)det(B) = 1, so it is not a group at all.

@fresh_42 that is it??

This is true.

But I had a similar problem as @jambaugh since I didn't understand your wording.