# Lemma of Morse in dimension 1

I want to write the Morse lemma which is in dimension $n$ :

Let $p$ be a non-degenerate critical point for $f$.

Then there is a local coordinate system $(y^1,...,y^n)$ in a neighborhood $U$ of $p$ with $y^i(p) = 0$ for all $i$ and such that the identity $f= f(p) - (y^1)^2- ... -(y^{\lambda})^ 2 + (y^{\lambda +1})^2+...+(y^n)^2$ holds throughout $U$, where $\lambda$ is the index of $f$ at $p$.

into dimension 1 .

but i don't know how ? , beacause i don't know hwo are the $y^i$ functions ?

thank you

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The $y^i$ are just the local coordinates. For dimension $1$, just take $n = 1$:
Let $M$ be a smooth $1$-manifold and $f: M \longrightarrow \Bbb R$ be a smooth function. Suppose $p$ is a non-degenerate critical point of $f$.
Then there exists a local coordinate system $(y^1)$ in a neighborhood $U \subset M$ of $p$ with $y^1(p) = 0$ satisfying the identity $$f(y^1) = f(p) - (y^1)^2$$ if the Morse index of $f$ at $p$ is $1$ and the identity $$f(y^1) = f(p) + (y^1)^2$$ if the Morse index of $f$ at $p$ is $0$.
@Vrouvrou: An atlas is a collection of local charts $(U_\alpha, \varphi_\alpha)$. So each $\varphi_\alpha$ is a homeomorphism $\varphi_\alpha: U_\alpha \longrightarrow \Bbb R^n$. So we can write $\varphi_\alpha$ in terms of its components $y^i$ in $\Bbb R^n$: $\varphi_\alpha(p) = (y^1(p), \dots, y^n(p))$. The $y^i$ are called the local coordinates for the chart $(U_\alpha, \varphi_\alpha)$. Note that $f(y^1, \dots, y^n)$ is really shorthand for $(f \circ \varphi_\alpha^{-1})(y^1, \dots, y^n)$, since $f$ is a function on $M$ but the $y^i$ map to $\Bbb R$. – Henry T. Horton Apr 2 '13 at 18:00
so ,i can say: Let f : M→R a Morse function and x ∈ M a critical point of f. then there existe a charte of M, $\psi : U\rightarrow R$, with $\psi(x) = 0$, such that $f \circ \psi^{−1} : R\rightarrow R$ soit de la forme $f \circ \psi^{−1}(x) = −x^2$ if index of $f$=1 , and $f \circ \psi^{−1}(x) = x^2$ if the index of $f=0$ – Vrouvrou Apr 2 '13 at 20:03
@Vrouvrou: That's almost right, but I would call the critical point something else, like $p \in M$, so that $\psi(p) = 0$. This is because $p$ is in the manifold but $x$ is in $\Bbb R$. – Henry T. Horton Apr 2 '13 at 20:31