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If I understood correctly, you are asking whether the spectral gap $\gamma=1-\rho$ of a non-reversible Markov chain $P$ provides any universal control on the Poincaré constant (which is the spectral gap of the additive reversibilization of $P$, or in your notation, $1-\|P\|$). The answer is no, even on finite state spaces: consider the Markov chain on $\{0,1\}^n$ which, at each step, replaces the current state $x=(x_1,\ldots,x_n)$ with either $(x_2,\ldots,x_{n-1},0)$ or $(x_2,\ldots,x_{n-1},1)$, each with probability a half. This chain has the maximum possible spectral gap, namely $\gamma=1$. Yet, its Poincaré constant tends to $0$ as $n\to\infty$.

If I understood correctly, you are asking whether the spectral gap $\gamma=1-\rho$ of a non-reversible Markov chain $P$ provides any universal control on the Poincaré constant (which is the spectral gap of the additive reversibilization of $P$, or in your notation, $1-\|P\|$). The answer is no, even on finite state spaces: consider the Markov chain on $\{0,1\}^n$ which, at each step, replaces the current state $x=(x_1,\ldots,x_n)$ with either $(x_2,\ldots,x_{n},0)$ or $(x_2,\ldots,x_{n},1)$, each with probability a half. This chain has the maximum possible spectral gap, namely $\gamma=1$. Yet, its Poincaré constant tends to $0$ as $n\to\infty$.

If I understood correctly, you are asking whether the spectral gap $\gamma=1-\rho$ of a non-reversible Markov chain provides any universal control on the Poincaré constant (which is the spectral gap of the additive reversibilization of the chain). The answer is no, even on finite state spaces: consider the Markov chain on $\{0,1\}^n$ which, at each step, replaces the current state $x=(x_1,\ldots,x_n)$ with either $(x_2,\ldots,x_{n-1},0)$ or $(x_2,\ldots,x_{n-1},1)$, each with probability a half. This chain has the maximum possible spectral gap, namely $\gamma=1$. Yet, its Poincaré constant tends to $0$ as $n\to\infty$.

If I understood correctly, you are asking whether the spectral gap $\gamma=1-\rho$ of a non-reversible Markov chain $P$ provides any universal control on the Poincaré constant (which is the spectral gap of the additive reversibilization of $P$, or in your notation, $1-\|P\|$). The answer is no, even on finite state spaces: consider the Markov chain on $\{0,1\}^n$ which, at each step, replaces the current state $x=(x_1,\ldots,x_n)$ with either $(x_2,\ldots,x_{n-1},0)$ or $(x_2,\ldots,x_{n-1},1)$, each with probability a half. This chain has the maximum possible spectral gap, namely $\gamma=1$. Yet, its Poincaré constant tends to $0$ as $n\to\infty$.

If I understood correctly, you are asking whether the spectral gap of a non-reversible Markov chain gives any general control on its Poincaré constant (which is the spectral gap of the additive reversibilization). The answer is no, even on finite state spaces: consider the Markov chain on $\{0,1\}^n$ which, at each step, replaces the current state $x=(x_1,\ldots,x_n)$ with either $(x_2,\ldots,x_{n-1},0)$ or $(x_2,\ldots,x_{n-1},1)$, each with probability a half. This chain has spectral radius $0$, i.e. spectral gap $1$. Yet, its Poincaré constant tends to $0$ as $n\to\infty$ (like $1/n^2$).

If I understood correctly, you are asking whether the spectral gap $\gamma=1-\rho$ of a non-reversible Markov chain provides any universal control on the Poincaré constant (which is the spectral gap of the additive reversibilization of the chain). The answer is no, even on finite state spaces: consider the Markov chain on $\{0,1\}^n$ which, at each step, replaces the current state $x=(x_1,\ldots,x_n)$ with either $(x_2,\ldots,x_{n-1},0)$ or $(x_2,\ldots,x_{n-1},1)$, each with probability a half. This chain has the maximum possible spectral gap, namely $\gamma=1$. Yet, its Poincaré constant tends to $0$ as $n\to\infty$.