Well, I need a deformation retract from $GL_n^{+}(\mathbb{R})$ to $SO(n)$

Here is what I tried, let $A\in GL_n^{+}(\mathbb{R})$ $A=(A_1,\dots,A_n)$ where $A_i$'s are column vectors, Recall that the Gram-Schmidt algorithm turns A into an orthogonal matrix by the following sequence of steps. First normalise $A_1$ (i.e. make it unit length) $A_1\mapsto \frac{A_1}{|A_1|}$ next I make $A_2$ orthogonal to $A_1$ like $A_2\mapsto A_2-\langle A_1,A_2\rangle A_1$ and normalize $A_2\mapsto \frac{A_2}{|A_2|}$ like this up to $A_n$

But I am not getting an explicit homotopy which gives me a deformation retract $GL_n^{+}(\mathbb{R})$ to $SO(n)$

  • 1
    $\begingroup$ $A_1 \mapsto \left[\left(1 - t\right) + \frac{t}{|A_1|}\right]A_1$, $A_2 \mapsto A_2 - t \langle A_1, A_2 \rangle A_1$, no? $\endgroup$ – Alexei Averchenko Oct 16 '12 at 12:20

By QR decomposition you can write any $A \in \textrm{GL}_n^+ (\Bbb{R})$ as an orthogonal matrix $Q$ times an upper triangular matrix $R$ which you can arrange to be such that $\det Q = 1$. Clearly you can homotope $R$ to the identity matrix and thus $\textrm{GL}_n^+(\Bbb{R})$ deformation retracts onto $\textrm{SO}(n)$.

If you want an explicit deformation retract, you can take the function $F : \textrm{GL}_n^+(\Bbb{R}) \times I \to \textrm{SO}(n)$ given as follows. Write any $A$ in the connected component of the identity as $A= QR$ with $R = (a_{ij})$ upper triangular with entries on the diagonal all positive. Then

$$F(A,t) = Q\times \left(\begin{array}{ccccc} (t) + (1-t)a_{11} & (1-t)a_{12} & \ldots & (1-t)a_{1n} \\ 0 & (t) + (1-t)a_{22} & \ldots & (1-t) a_{2n} \\ & \ddots& \vdots \\ &&\ddots \\ && &t + (1-t)a_{nn} \end{array}\right)$$

is the required deformation retract.

You should note that the same proof shows that $\textrm{GL}_n(\Bbb{C})$ deformation retracts onto $\textrm{U}(n)$.


Here is a geometric way to see this. To any ordered basis $(v_1,v_2,\ldots,v_n)$ of your vector space $V$ associate the "flag" of subspaces $V_0=\{0\}$, $V_1=\langle v_1\rangle$, $V_2=\langle v_1,v_2\rangle$, ... $V_n=\langle v_1,v_2,\ldots,v_n\rangle=V$. The Gram-Schmidt algorithm turns any such basis into an orthonormal basis $(b_1,\ldots,b_n)$ that gives rise to the same flag of subspaces. It is moreover the unique such basis (orthonomal and with the same flag) for which in addition each $b_i$, inside $V_i$, is on the same side of the hyperplane $V_{i-1}$ as the original basis vector $v_i$.

Now taking $V=\Bbb R^n$ we can identify $GL_n^+(\Bbb R)$ with the set of ordered bases $(v_1,v_2,\ldots,v_n)$ with $\det(v_1,v_2,\ldots,v_n)>0$, and $SO(n)$ with the set of ordered orthonormal bases $(b_1,b_2,\ldots,b_n)$ with $\det(b_1,b_2,\ldots,b_n)>0$. Now for such a basis $(v_1,v_2,\ldots,v_n)$ let $(b_1,\ldots,b_n)$ be the orthonormal basis associated to it under Gram-Schmidt, and simultaneously (or successively if you prefer) deform every $v_i$ linearly to $b_i$, as $t\mapsto (1-t)v_i+tb_i$. The intermediate vectors stay inside $V_i$, and since $b_i$ is on the same side as $v_i$, they never enter $V_{i-1}$. This means the deformed vectors stay linearly independent at all times, so the deformation takes place inside $GL_n(\Bbb R)$. As the determinant cannot vanish anywhere we have $\det(v_1,v_2,\ldots,v_n)>0\implies \det(b_1,b_2,\ldots,b_n)>0$ and we have a deformation retract of $GL_n^+(\Bbb R)$ to $SO(n)$. It is in fact a strong deformation retract: elements of $SO(n)$ remain fixed.

  • 2
    $\begingroup$ fabulas explaination dear sir $\endgroup$ – Marso Oct 17 '12 at 5:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.