# Shear transformations

The Wikipedia article is not enough, and sadly it does not have any references.

I understand they are linear transformations. Do they form a group? How do they look like for n-dimensional vectors? How many independent shears can I do in n-dimensions? Are they related to the orthogonal group SO(D), perhaps contained?

Google takes me to this other website. It is almost enough, but again no references.

I am trying to do some physics with this, so I do not think you will tolerate my explanation. ;-)

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What do you mean by SO(2,D)? I would usually interpret it as the group of linear transformations preserving a quadratic form of signature (2,D), but then the answer is clearly no, e.g., set D = 0 and notice that angles are not preserved in the picture on that Wikipedia page. – Reid Barton Nov 4 '09 at 16:19
Yes, I meant that group. Yes, angles are not preserved so no relation to conformal transformations. Thanks! Still, any idea where I can get a more formal treatment of these type of linear transformations? – M. E. Irizarry-Gelpí Nov 4 '09 at 20:36
"Shear transformations" isn't really a subject, so you won't find a book with that title. Could you please clarify what it is that you want to know about shears, or what you're trying to do with them? As stated, I don't think this question has an answer. – Anton Geraschenko Nov 4 '09 at 21:56
I edited the question. – M. E. Irizarry-Gelpí Nov 4 '09 at 22:54
No, they don't form a group. However you define a shear, any upper (or lower) triangular matrix with 1s on the diagonal is a composition of shears. But any determinant 1 matrix can be written as a product of such a lower and upper triangular matrix, and it's certainly not true that any determinant 1 matrix is a shear (for example, no non-identity element of SO(n) is a shear, since non-trivial shears always change some angles). You should really just say what you're trying to do; many math people do lots of physics, and this question is pretty impossible to answer without any motivation. – Anton Geraschenko Nov 5 '09 at 2:22

It might be helpful to note that, in two dimensions, a shear transformation is exactly one whose Jordan canonical form is $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ or $\begin{pmatrix}1&0\\0&1\end{pmatrix}$ if you include the identity as a shear transformation. These are exactly the transformations whose eigenvalues are all $1$, so-called unipotent matrices. The unipotent $2 \times 2$ matrices do not form a group: $\begin{pmatrix}1&1\\0&1\end{pmatrix}\times\begin{pmatrix}1&0\\1&1\end{pmatrix}$ is not unipotent.