# What does \$Bbb S^{n-1}times mathbb{R}\$ stand for?

Let \$ngeqslant 1,f: Bbb R^n – { 0 } to S^{n – 1} times Bbb R,xto(frac{x}{||x||},ln(||x||))\$ is a homeomorphism which the inverse is \$f^{-1}:S^{n-1}timesmathbb{R}tomathbb{R}^n-{0},(y,t)to e^{t}y\$.

I was reading this example but I cannot understand what \$S^{n-1}timesmathbb{R}\$ means. I suspect that \$S^{n-1}\$ is the counter dominion of the function. y

Questions:

1) Is \$S^{n-1}\$ the counter-dominion of f?

2) Why is the counter dominion \$S^{n-1}\$? Where does the \$n-1\$ comes from?

Thanks in advance!

## Answer

Answer \$Bbb S^{n-1}\$ is the unit sphere in \$Bbb R^n\$ ie
\$\$Bbb S^{n-1} ={xiinBbb R^n: |xi|=1}\$\$
\$n-1\$ represent the DIMENSION OF \$Bbb S^{n-1}\$ as a MANIFOLD
precisely,
\$\$dim Bbb S^{n-1}= n-1, ~~~~~ dim Bbb R^{n} =n\$\$

For instance, in dimension 2, i.e in \$Bbb R^{2}\$
the unit circle is defined
\$\$Bbb S^{1} ={xiinBbb R^2: |xi|=1} equiv {e^{itheta}: thetain[0,2pi)}\$\$
is of dimension one. \$dimBbb S^1 =1\$ roughly speaking you see \$Bbb S^1\$ as the real line \$Bbb R\$.