For given Pontryagin space the norm induced topology is not depending on the fundamental decomposition For given Hilbert space $H$ consider its corresponding antispace, i.e. the vector space endowed with its negative inner product. $\Pi$ is now called a Pontryagin space if it can be written as a direct sum
\begin{equation}
    \Pi=N \oplus N^{\perp}
\end{equation}
where $\left(N^{\perp},\langle\cdot, \cdot\rangle_{N \perp}\right)$ is a Hilbert space and $\left(N,\langle\cdot, \cdot\rangle_{N}\right)$  is the antispace of a finite-dimensional Hilbert space. The space $\Pi$ is equipped with an indefinite inner product the so called fundamental decomposition
\begin{equation}
(v, w)=\left\langle v_{+}, w_{+}\right\rangle_{N^{\perp}}+\left\langle v_{-}, w_{-}\right\rangle_{N}, \quad v_{+}, w_{+} \in N^{\perp}, v_{-}, w_{-} \in N
\end{equation}
The supspace $N$ denotes the negative subspace of $\Pi$, since from the definition we immediately get that $(v, v)<0$ for all $v \in N \backslash\{0\} .$
The subspace $N$ is not unique. Nevertheless the dimension is unique and is called the index of the Pontryagin space.
For a given Pontryagin space $(\Pi, (\cdot,\cdot))$ its induced Hilbert space is naturally given via the fundamental decomposition
\begin{equation}
    [v, w]=\left(v_{+}, w_{+}\right)-\left(v_{-}, w_{-}\right), \quad v_{+}, w_{+} \in N^{\perp}, v_{-}, w_{-} \in N
\end{equation}
So why are for two different fundamental decompositions the norms equivalent and thus the topologies are the same?
 A: I have found the following:
Proof. Let $N_{1} \oplus N_{1}^{\perp}$ and $N_{2} \oplus N_{2}^{\perp}$ be two different fundamental decompositions of our Pontryagin space $(\Pi,(\cdot, \cdot))$. Let's denote by $\|\cdot\|_{1},\|\cdot\|_{2}$ the corresponding norms by the induced scalar products. Consider the identical mapping id : $\left\langle\Pi,\|\cdot\|_{1}\right\rangle \rightarrow\left\langle\Pi,\|\cdot\|_{2}\right\rangle .$ Let $\left(x_{n}\right)_{n \in \mathbb{N}}$ be a sequence in $\Pi$, where $x_{n} \stackrel{\|\cdot\|_{1}}{\longrightarrow} x, x_{n} \stackrel{\|-\|_{2}}{\longrightarrow} x^{\prime} .$ It holds
$$
\left[x_{n}, y\right] \rightarrow[x, y],\left[x_{n}, y\right] \rightarrow\left[x^{\prime}, y\right], \quad y \in \Pi
$$
and thus $x=x^{\prime}$. Now we know by the closed graph theorem that id is continuous and by the open mapping theorem we obtain a homeomorphism, $\mathcal{T}\left(\|\cdot\|_{1}\right)=\mathcal{T}\left(\|\cdot\|_{2}\right)$.
Please let me know if you see a mistake.
Best regards.
