T-5 - Revision history

Ros: /* To know more */

2026-03-11T10:26:38Z

To know more

← Older revision		Revision as of 12:26, 11 March 2026
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	== To know more ==		== To know more ==
	In Problem 5, we have defined the gradient <math>{\nabla}_\perp E(\vec{\sigma})</math> and the Hessian <math>{\nabla}^2_\perp E(\vec{\sigma})</math> of the function <math> E(\vec{\sigma})</math> restricted to a manifold (the sphere). These are called Riemannian gradient and Hessian. How are they defined?		In Problem 5, we have defined the gradient <math>{\nabla}_\perp E(\vec{\sigma})</math> and the Hessian <math>{\nabla}^2_\perp E(\vec{\sigma})</math> of the function <math> E(\vec{\sigma})</math> restricted to a manifold (the sphere). These are called Riemannian gradient and Hessian. How are they defined?
			First, a few notions of geometry: we define with <math> \hat \Pi(\vec{\sigma}) </math> the projector on the tangent plane to the sphere at <math> \vec{\sigma}</math>: this is the plane orthogonal to the vector <math> \vec{\sigma}</math>. The gradient <math> \nabla_\perp E(\vec{\sigma}) </math> is a <math> (N-1)</math>-dimensional vector that is obtained projecting the gradient <math> [\nabla E(\vec{\sigma})]_i=\partial E/\partial \sigma_i </math> on the tangent plane, <math> \nabla_\perp E(\vec{\sigma})=\hat \Pi \nabla E(\vec{\sigma})</math>. The Hessian <math> {\nabla}^2_\perp E(\vec{\sigma}) </math> is a <math> (N-1) \times (N-1)</math>-dimensional matrix that is obtained from the Hessian <math> [\nabla^2 E(\vec{\sigma})]_{ij}=\partial^2 E/\partial \sigma_i \partial \sigma_j </math> as <math> {\nabla}^2_\perp E(\vec{\sigma})= \hat \Pi(\vec{\sigma}) \, \nabla^2 E(\vec{\sigma}) \, \hat\Pi(\vec{\sigma}) - N^{-1}\nabla E(\vec{\sigma}) \cdot \vec{\sigma} \mathbb{I}</math> where <math> \mathbb{I}</math> is the identity matrix.
			For a discussion on this (with pictures), have a look to B3, page 13 in the notes:
	First, a few notions of geometry: we define with <math> \hat \Pi(\vec{\sigma}) </math> the projector on the tangent plane to the sphere at <math> \vec{\sigma}</math>: this is the plane orthogonal to the vector <math> \vec{\sigma}</math>. The gradient <math> \nabla_\perp E(\vec{\sigma}) </math> is a <math> (N-1)</math>-dimensional vector that is obtained projecting the gradient <math> [\nabla E(\vec{\sigma})]_i=\partial E/\partial \sigma_i </math> on the tangent plane, <math> \nabla_\perp E(\vec{\sigma})=\hat \Pi \nabla E(\vec{\sigma})</math>. The Hessian <math> \nabla^2_\perp E(\vec{\sigma}) </math> is a <math> (N-1) \times (N-1)</math>-dimensional matrix that is obtained from the Hessian <math> [\nabla^2 E(\vec{\sigma})]_{ij}=\partial^2 E/\partial \sigma_i \partial \sigma_j </math> as <math> \nabla^2_\perp E(\vec{\sigma})= \hat \Pi(\vec{\sigma}) \, \nabla^2 E(\vec{\sigma}) \, \hat\Pi(\vec{\sigma}) - N^{-1}\nabla E(\vec{\sigma}) \cdot \vec{\sigma} \mathbb{I}</math> where <math> \mathbb{I}</math> is the identity matrix.		* Ros. High-dimensional random landscapes: From typical to large deviations [https://scipost.org/SciPostPhysLectNotes.102/pdf]




	* ~~Castellani, Cavagna~~. ~~Spin~~-~~Glass Theory for Pedestrians~~ [https://~~arxiv~~.org/~~abs/cond-mat~~/~~0505032~~]

Ros: /* To know more */

2026-03-11T10:24:06Z

To know more

← Older revision		Revision as of 12:24, 11 March 2026
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	== To know more ==		== To know more ==
	In Problem 5, we have defined		In Problem 5, we have defined the gradient <math>{\nabla}_\perp E(\vec{\sigma})</math> and the Hessian <math>{\nabla}^2_\perp E(\vec{\sigma})</math> of the function <math> E(\vec{\sigma})</math> restricted to a manifold (the sphere). These are called Riemannian gradient and Hessian. How are they defined?


			First, a few notions of geometry: we define with <math> \hat \Pi(\vec{\sigma}) </math> the projector on the tangent plane to the sphere at <math> \vec{\sigma}</math>: this is the plane orthogonal to the vector <math> \vec{\sigma}</math>. The gradient <math> \nabla_\perp E(\vec{\sigma}) </math> is a <math> (N-1)</math>-dimensional vector that is obtained projecting the gradient <math> [\nabla E(\vec{\sigma})]_i=\partial E/\partial \sigma_i </math> on the tangent plane, <math> \nabla_\perp E(\vec{\sigma})=\hat \Pi \nabla E(\vec{\sigma})</math>. The Hessian <math> \nabla^2_\perp E(\vec{\sigma}) </math> is a <math> (N-1) \times (N-1)</math>-dimensional matrix that is obtained from the Hessian <math> [\nabla^2 E(\vec{\sigma})]_{ij}=\partial^2 E/\partial \sigma_i \partial \sigma_j </math> as <math> \nabla^2_\perp E(\vec{\sigma})= \hat \Pi(\vec{\sigma}) \, \nabla^2 E(\vec{\sigma}) \, \hat\Pi(\vec{\sigma}) - N^{-1}\nabla E(\vec{\sigma}) \cdot \vec{\sigma} \mathbb{I}</math> where <math> \mathbb{I}</math> is the identity matrix.

	* Castellani, Cavagna. Spin-Glass Theory for Pedestrians [https://arxiv.org/abs/cond-mat/0505032]



			* Castellani, Cavagna. Spin-Glass Theory for Pedestrians [https://arxiv.org/abs/cond-mat/0505032]
	~~<!----------OLD---------~~
	~~First~~, a few notions of geometry: we define with <math> \hat \Pi(\vec{\sigma}) </math> the projector on the tangent plane to the sphere at <math> \vec{\sigma}</math>: this is the plane orthogonal to the vector <math> \vec{\sigma}</math>. ~~The gradient <math> \nabla_\perp E(\vec{\sigma}) </math> is a <math> (N-1)</math>~~-~~dimensional vector that is obtained projecting the gradient <math>~~ [~~\nabla E(\vec{\sigma})]_i=\partial E~~/~~\partial \sigma_i <~~/~~math> on the tangent plane, <math> \nabla_\perp E(\vec{\sigma})=\hat \Pi \nabla E(\vec{\sigma})<~~/~~math>. The Hessian <math> \nabla^2_\perp E(\vec{\sigma}) <~~/~~math> is a <math> (N-1) \times (N~~-~~1)<~~/~~math>-dimensional matrix that is obtained from the Hessian <math> [\nabla^2 E(\vec{\sigma})~~]_{ij}=\partial^2 E/\partial \sigma_i \partial \sigma_j </math> as <math> \nabla^2_\perp E(\vec{\sigma})= \hat \Pi(\vec{\sigma}) \, \nabla^2 E(\vec{\sigma}) \, \hat\Pi(\vec{\sigma}) - N^{-1}\nabla E(\vec{\sigma}) \cdot \vec{\sigma} \mathbb{I}</math> where <math> \mathbb{I}</math> is the identity matrix. <br>-->

Ros: /* Check out: key concepts */

2026-03-11T10:22:20Z

Check out: key concepts

← Older revision		Revision as of 12:22, 11 March 2026
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	Gradient descent, rugged landscapes, metastable states, landscape’s complexity.		Gradient descent, rugged landscapes, metastable states, landscape’s complexity.


	== To know more ==		== To know more ==

Ros: /* Check out: key concepts */

2026-03-11T10:21:50Z

Check out: key concepts

← Older revision	Revision as of 12:21, 11 March 2026
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		== To know more ==
		In Problem 5, we have defined



		* Castellani, Cavagna. Spin-Glass Theory for Pedestrians [https://arxiv.org/abs/cond-mat/0505032]

Ros: /* Problem 5: the Kac-Rice formula and the complexity */

2026-03-09T08:57:59Z

Problem 5: the Kac-Rice formula and the complexity

← Older revision		Revision as of 10:57, 9 March 2026
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	<li><em> Gaussianity and correlations.</em> Show that:		<li><em> Gaussianity and correlations.</em> Show that:
	<math display="block">		<math display="block">
	\mathbb{E}[\mathcal{N}(\epsilon)]= (2 \pi e)^{\frac{N}{2}} \, \frac{1}{(2 \pi \, p)^{\frac{N-1}{2}}}\; \sqrt{\frac{N}{2 \pi}} e^{-\frac{N \epsilon^2}{2}}\;\~~overline~~{\|\text{det} \left( M- p \epsilon \mathbb{I} \right)\|}		\mathbb{E}[\mathcal{N}(\epsilon)]= (2 \pi e)^{\frac{N}{2}} \, \frac{1}{(2 \pi \, p)^{\frac{N-1}{2}}}\; \sqrt{\frac{N}{2 \pi}} e^{-\frac{N \epsilon^2}{2}}\;\mathbb{E}[\|\text{det} \left( M- p \epsilon \mathbb{I} \right)\|]
	</math>		</math>
	by using the following facts:		by using the following facts:
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	where the matrix <math> M </math> has random entries with zero average and correlations		where the matrix <math> M </math> has random entries with zero average and correlations
	<math>		<math>
	\~~overline~~{{M}_{\alpha \beta} \, {M}_{\gamma \delta}}= \frac{p (p-1)}{ N} \left( \delta_{\alpha \gamma} \delta_{\beta \delta}+ \delta_{\alpha \delta} \delta_{\beta \gamma}\right).		\mathbb{E}[{M}_{\alpha \beta} \, {M}_{\gamma \delta}]= \frac{p (p-1)}{ N} \left( \delta_{\alpha \gamma} \delta_{\beta \delta}+ \delta_{\alpha \delta} \delta_{\beta \gamma}\right).
	</math>		</math>
	</ul>		</ul>

Ros: /* Problem 5: the Kac-Rice formula and the complexity */

2026-03-09T08:42:03Z

Problem 5: the Kac-Rice formula and the complexity

← Older revision		Revision as of 10:42, 9 March 2026
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	<li><em> Statistical rotational invariance.</em> If you solved <code>Exercise 9</code>, you have shown that the correlations of the energy landscape of the <math>p</math>-spin model are <math>\mathbb{E}[E(\vec{\sigma}) E(\vec{\tau})]= N q(\vec{\sigma}, \vec{\tau})^p</math>, where <math>q</math> is the overlap: in which sense the correlation function is rotationally invariant? Justify why rotational invariance implies that		<li><em> Statistical rotational invariance.</em> If you solved <code>Exercise 9</code>, you have shown that the correlations of the energy landscape of the <math>p</math>-spin model are <math>\mathbb{E}[E(\vec{\sigma}) E(\vec{\tau})]= N q(\vec{\sigma}, \vec{\tau})^p</math>, where <math>q</math> is the overlap: in which sense the correlation function is rotationally invariant? Justify why rotational invariance implies that
	<math display="block">		<math display="block">
	\mathbb{E}[\mathcal{N}(\epsilon)]= (2 \pi e)^{\frac{N}{2}} \, p_{\epsilon}(\vec{1})~~e^{o(N)}~~		\mathbb{E}[\mathcal{N}(\epsilon)]= (2 \pi e)^{\frac{N}{2}}e^{o(N)} \, p_{\epsilon}(\vec{1})
	</math>		</math>
	where <math> \vec{1}=(1,1,1, \cdots, 1) </math> is one fixed vector belonging to the surface of the sphere. Where does the prefactor arise from?		where <math> \vec{1}=(1,1,1, \cdots, 1) </math> is one fixed vector belonging to the surface of the sphere. Where does the prefactor arise from?

Ros: /* Problem 5: the Kac-Rice formula and the complexity */

2026-03-09T08:41:38Z

Problem 5: the Kac-Rice formula and the complexity

← Older revision		Revision as of 10:41, 9 March 2026
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	<li><em> Statistical rotational invariance.</em> If you solved <code>Exercise 9</code>, you have shown that the correlations of the energy landscape of the <math>p</math>-spin model are <math>\mathbb{E}[E(\vec{\sigma}) E(\vec{\tau})]= N q(\vec{\sigma}, \vec{\tau})^p</math>, where <math>q</math> is the overlap: in which sense the correlation function is rotationally invariant? Justify why rotational invariance implies that		<li><em> Statistical rotational invariance.</em> If you solved <code>Exercise 9</code>, you have shown that the correlations of the energy landscape of the <math>p</math>-spin model are <math>\mathbb{E}[E(\vec{\sigma}) E(\vec{\tau})]= N q(\vec{\sigma}, \vec{\tau})^p</math>, where <math>q</math> is the overlap: in which sense the correlation function is rotationally invariant? Justify why rotational invariance implies that
	<math display="block">		<math display="block">
	\mathbb{E}[\mathcal{N}(\epsilon)]= (2 \pi e)^{\frac{N}{2}} \, p_{\epsilon}(\vec{1})		\mathbb{E}[\mathcal{N}(\epsilon)]= (2 \pi e)^{\frac{N}{2}} \, p_{\epsilon}(\vec{1})e^{o(N)}
	</math>		</math>
	where <math> \vec{1}=(1,1,1, \cdots, 1) </math> is one fixed vector belonging to the surface of the sphere. Where does the prefactor arise from?		where <math> \vec{1}=(1,1,1, \cdots, 1) </math> is one fixed vector belonging to the surface of the sphere. Where does the prefactor arise from?

Ros: /* Dynamics, optimization, trapping local minima */

2026-02-14T18:15:52Z

Dynamics, optimization, trapping local minima

← Older revision		Revision as of 20:15, 14 February 2026
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	</ul>		</ul>
	<br>		<br>



	<div style="font-size:89%">		<div style="font-size:89%">

Ros: /* Problem 5: the Kac-Rice formula and the complexity */

2026-02-08T19:17:58Z

Problem 5: the Kac-Rice formula and the complexity

← Older revision		Revision as of 21:17, 8 February 2026
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	<ol start="2">		<ol start="2">
	<li><em> Statistical rotational invariance.</em> If you solved Exercise 9, you have shown that the correlations of the energy landscape of the <math>p</math>-spin model are <math>\mathbb{E}[E(\vec{\sigma}) E(\vec{\tau})]= N q(\vec{\sigma}, \vec{\tau})^p</math>, where <math>q</math> is the overlap: in which sense the correlation function is rotationally invariant? Justify why rotational invariance implies that		<li><em> Statistical rotational invariance.</em> If you solved <code>Exercise 9</code>, you have shown that the correlations of the energy landscape of the <math>p</math>-spin model are <math>\mathbb{E}[E(\vec{\sigma}) E(\vec{\tau})]= N q(\vec{\sigma}, \vec{\tau})^p</math>, where <math>q</math> is the overlap: in which sense the correlation function is rotationally invariant? Justify why rotational invariance implies that
	<math display="block">		<math display="block">
	\mathbb{E}[\mathcal{N}(\epsilon)]= (2 \pi e)^{\frac{N}{2}} \, p_{\epsilon}(\vec{1})		\mathbb{E}[\mathcal{N}(\epsilon)]= (2 \pi e)^{\frac{N}{2}} \, p_{\epsilon}(\vec{1})

Ros: /* Problem 5: the Kac-Rice formula and the complexity */

2026-02-08T19:17:02Z

Problem 5: the Kac-Rice formula and the complexity

← Older revision		Revision as of 21:17, 8 February 2026
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	=== Problem 5: the Kac-Rice formula and the complexity ===		=== Problem 5: the Kac-Rice formula and the complexity ===