Invariant framing of $S^1$ and "the other one" It is known that the left invariant framing of $S^1$ generates $\Omega_1^{fr}$ (the stably framed bordism group of 1-dimensional manifolds). I do not understand how "the other" framing looks like which gives the zero element in $\Omega_1^{fr}$. Can someone explain this to me?
(First I thought it would be the framing obtained by reversing orientation, but this should give the same element in $\Omega_1^{fr}$.)
 A: Non-stably the zero element is easy to see i think. just imagine a circle in $\mathbb{R}^2$ with an outward (or inward) pointing normal. You can see by hand that this is framed cobordant to the empty set. But now you can just suspend: Take the same circle with this framing and see it sitting inside $\mathbb{R}^2\times\mathbb{R}^2$ with a constant basis of the normal space in the other two directions. Thus the framing of the standard circle $\{(x,y,0,0)\in \mathbb{R}^4\,x^2+y^2=1\}$ at the point $(x,y,0,0)$ is $e_1=(x,y,0,0),e_2=(0,0,1,0),e_3=(0,0,0,1)$.
The non-zero element in framed cobordism can be seen as follows: take a non-zero contractible loop $S^1\rightarrow SO(3)$ and let it act on the basis above pointwise. 
EDIT: Consider the circle $S^1\subset \mathbb R^2$ with the outward pointing normal as a framing. Consider the $W=\{(x,y,z)\subset \mathbb R^2\times [0,1]\,|\,x^2+y^2+2z^2=1.\}$ with framing $(x,y,2z)$. This is a framed nullbordism. I encourage you to make a drawing. You can even go one dimension lower: Take $S^0\subset \mathbb R$ (two points) with outward normal. This extends over an arc $\gamma\subset \mathbb R\times[0,1]$ with boundary on $S^0\times\{0\}$. 
